The time required for one half-life in a first-order reaction is given by the formula below and it shows the result how it will take 108 minutes for 4 half-lives to occur in this first-order reaction.
The formula is t1/2 = 0.693 / k
where k is the rate constant for the reaction.
Given that the half-life of the reaction is 27 minutes, we can rearrange this formula to solve for the rate constant:
k = 0.693 / t1/2
k = 0.693 / 27 min
k = 0.0257 min^-1
Now we can use the first-order rate law to determine the time required for 4 half-lives to occur:
t = (4 x t1/2) = (4 x 27 min) = 108 min
So, it takes 108 mins that is equal to one hour forty eight minutes to occur four half lives.
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At 25 °C, an aqueous solution has an equilibrium concentration of 0.00343M for a generic cation, A+(aq), and 0.00343M for a generic anion, B−(aq). What is the equilibrium constant, sp, of the generic salt AB(s)?
[tex] \:\:\:\:\:\:\star [/tex]For the general solubility equilibrium [tex]\sf \underline{AB \longrightarrow A^+ + B^-} [/tex]the solubility product has the following expression-
[tex] \:\:\:\:\:\:\star\longrightarrow \sf\underline{K_{(sp)} = [A^+] \times [B^-]}\\[/tex]
As per question, we are given that-
Equilibrium concentration for generic cation,[tex]\sf [A^+][/tex]= 0.00343MEquilibrium concentration for generic anion, [tex]\sf [B^-] [/tex]= 0.00343M[tex] \:\:\:\:\:\:\star [/tex] Now that we have all the required values, so we can substitute these values into the Ksp expression and solve for Ksp-
[tex] \:\:\:\:\:\:\star\longrightarrow \sf\underline{K_{(sp)} = [A^+] \times [B^-]}\\[/tex]
[tex] \:\:\:\:\:\:\longrightarrow \sf K_{(sp)} = 0.00343 \:M\times 0.00343\:M\\[/tex]
[tex] \:\:\:\:\:\:\longrightarrow \sf K_{(sp)} = 0.00343 \:molL^{-1}\times 0.00343\:molL^{-1}\\[/tex]
[tex] \:\:\:\:\:\:\longrightarrow \sf \underline{K_{(sp)} = 1.17649\times 10^{-5} \: mol^2L^{-2}}\\[/tex]
Hence, the equilibrium constant(Ksp) of the generic salt AB(s) is [tex]\sf\underline{\boxed{\sf1.17649\times 10^{-5} \: mol^2L^{-2}}}.\\[/tex]Calculate AH for the reaction:
C2H4 (g) + 3O2 (g)--> 2H2O (g) + 2CO2 (g)
The enthalpy change or AH for the reaction is -1560.9 kJ/mol. This negative value indicates that the reaction is exothermic, meaning it releases energy in the form of heat to the surroundings.
To calculate the enthalpy change (ΔH) or AH for the reaction:C2H4 (g) + 3O2 (g) → 2H2O (g) + 2CO2 (g)
We can use the standard enthalpies of formation (∆Hf°) of the products and reactants to determine the overall enthalpy change. The ∆Hf° values can be found in a reference table or online database.
The balanced equation tells us that 1 mole of C2H4 reacts with 3 moles of O2 to form 2 moles of H2O and 2 moles of CO2. So, we can write:
ΔH° = [2∆Hf°(H2O) + 2∆Hf°(CO2)] - [∆Hf°(C2H4) + 3∆Hf°(O2)]
Substituting the ∆Hf° values for each substance, we get:
ΔH° = [2(-241.8 kJ/mol) + 2(-393.5 kJ/mol)] - [(52.3 kJ/mol) + 3(0 kJ/mol)] ΔH° = -1560.9 kJ/mol
Therefore, the enthalpy change or AH for the reaction is -1560.9 kJ/mol. This negative value indicates that the reaction is exothermic, meaning it releases energy in the form of heat to the surroundings.
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which mixture will not result in a neutral solution? select the correct answer below: 1 m naoh and 1 m hcl 1 m nh3 and 1 m hcl 1 m koh and 1 m hbr 1 m naoh and 1 m hi
When a strong acid and a strong base are mixed in equal amounts, they undergo a neutralization reaction, resulting in the formation of water and a salt.
Therefore, the mixture of 1 M NaOH and 1 M HCl will not result in a neutral solution but instead will form sodium chloride and water. On the other hand, the mixture of 1 M NH3 and 1 M HCl, and the mixture of 1 M KOH and 1 M HBr, will also undergo neutralization reactions but will result in the formation of ammonium chloride and potassium bromide, respectively. The mixture of 1 M NaOH and 1 M HI will also not result in a neutral solution but will form sodium iodide and water due to the reaction between the strong base NaOH and the weak acid HI.
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if there is no diethyl ether in the lab, what other solvent can you use as an alternative? select one: methanol ethyl acetate tetrahydrofuran water
While considering a good solvent alternative for diethyl ether, the best one will be ethyl acetate. It can be used for extraction due to its polarity and less toxicity.
Diethyl ether is one of the commonly used solvent in extraction process of non-polar or slightly polar organic compounds. This is because it does not have hydrogen bonding. So here methanol cannot be used as it has extensive hydrogen bonding and non-polar compounds might not dissolve.
Water also cannot be used because of its polar nature. So organic compounds does not dissolve. Tetrahydrofuran can be used as a solvent, but toxicity levels are higher compared to diethyl ether.
So the alternative that can be used is ethyl acetate, which is also widely used solvent in extraction of non-polar compounds. Also it has less toxicity compared to THF.
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a solution of pyridinium bromide has a ph of 3.10. what is the concentration of the pyridinium cation at equilibrium, in units of molarity?
The concentration of the pyridinium cation at equilibrium is 3.96 x 10^-9 M.Assuming that pyridinium bromide is a weak acid, we can use the acid dissociation constant (Ka) to calculate the concentration of the pyridinium cation at equilibrium. The Ka for pyridinium bromide is 5.6 x 10^-6.
Using the expression for Ka, we have:
Ka = [H+][C5H5NH+] / [C5H5NHBr]
At equilibrium, [H+] = 10^-pH = 10^-3.10 = 7.94 x 10^-4 M
Substituting the values into the equation and solving for [C5H5NH+], we get:
[C5H5NH+] = Ka * [C5H5NHBr] / [H+] = 5.6 x 10^-6 * [C5H5NHBr] / 7.94 x 10^-4 = 3.96 x 10^-9 M
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b) explain the meaning of the term ld50 (ed50). what is the ld50 concentration of cuso4 for brine shrimp?
The LD50 concentration of CuSO₄ for brine shrimp is reported to be around 2.75 ppm (parts per million), which means that if 50% of the brine shrimp population were exposed to this concentration of CuSO₄, they would die as a result of the exposure
The LD50 and ED50 are both terms commonly used in toxicology to express the effectiveness or toxicity of a substance.
The LD50, which stands for "lethal dose 50," is the amount of a substance required to cause death in 50% of the test population. It is typically expressed in units of milligrams or micrograms of the substance per kilogram of body weight of the test animal.
On the other hand, the ED50 stands for "effective dose 50," which is the amount of a substance required to produce a desired effect in 50% of the test population. It is commonly used in pharmacology to measure the potency of a drug.
In the case of the brine shrimp, the LD50 concentration of CuSO4 (copper sulfate) would be the amount of CuSO4 that would cause the death of 50% of the shrimp population in a given test. It is important to note that the LD50 can vary depending on various factors such as the species being tested, the method of exposure, and the duration of exposure.
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which two statements below are true of the reaction above? choose the 2 correct statements. a in the forward reaction, nitrogen and hydrogen combine to form ammonia. b in the forward reaction, ammonia decomposes into nitrogen and hydrogen. c in the reverse reaction, nitrogen and hydrogen combine to form ammonia. d in the reverse reaction, ammonia decomposes into nitrogen and hydrogen.
The two correct statements are:
A) In the forward reaction, nitrogen and hydrogen combine to form ammonia.
C) In the reverse reaction, nitrogen and hydrogen combine to form ammonia.
The given reaction is the synthesis of ammonia, which is represented as:
N2 + 3H2 → 2NH3
In the forward reaction, nitrogen and hydrogen combine to form ammonia, as stated in option A.
In the reverse reaction, ammonia decomposes back into nitrogen and hydrogen, which is the opposite of the forward reaction. The reverse reaction is represented as:
2NH3 → N2 + 3H2
Option B is incorrect because it describes the reverse reaction instead of the forward reaction. Option D is incorrect because it describes the decomposition of ammonia instead of the synthesis of ammonia.
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in the case of anionic polymerization using organometallic initiator, abby was able to get polystyrene with a number average molecular weight of 31,200 g/mole. please estimate the outcome of the new polymer if she double the initial monomer concentration and quadruple the initiator concentration from the original settings with all the other conditions/parameters and conversion being the same?
If Abby doubles the initial monomer concentration and quadruples the initiator concentration from the original settings, the outcome of the new polymer will likely have a higher number average molecular weight.
Abby's outcome of the new polymer will likely have a higher number average molecular weight because increasing the monomer concentration will lead to more monomer units being available for polymerization, while increasing the initiator concentration will lead to more initiation events, resulting in a higher degree of polymerization.
However, the exact molecular weight of the new polymer will depend on the efficiency of the polymerization reaction and any potential side reactions or termination events that may occur. It is possible that increasing the initiator concentration could also lead to more side reactions, such as chain transfer or termination, which could affect the final molecular weight distribution.
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How do I solve this?
As a result, the theoretical yield of Fe3O4 in moles is 1.426 x 105 molewhile the actual yield is 2.41 mole and the percent yield is 0.107%.
The reaction's chemically balanced equation is as follows:
4H₂ + Fe₃O₄ = 3Fe + 4H₂O
Fe₃O₄has a molar mass of 231.5326 g/mol7.
We need to know the mass of Fe₃O₄ created in order to calculate the real amount of Fe₃O₄ in moles. 559 g1 of Fe₃O₄ were produced.
We must first determine the quantity of Fe utilized in the reaction in order to compute the theoretical yield of magnitude Fe₃O₄ moles. 7.97 million g, or 7.97 x 106 g, of Fe are used1. Fe has a molar mass of 55.845 g/mol7. Hence, the amount of Fe utilized in the reaction was:
1.426 x 105 mole = (7.97 x 106 g) / (55.845 g/mole)
One mole of Fe creates one mole of Fe3O4 according to the chemical equation. As a result, 1.426 x 105 mole is the theoretical yield of magnitude Fe3O4 in moles.
In order to determine percent yield, we apply the formula:
(Actual yield / Theoretical yield) times 100% equals the percent yield.
Replacement of values
Yield is calculated as follows:
(559 g/(1.426 x 105 mole/231.5326 g/mol)) x 100% = **0.107%**6.
What common chemical formulas are there?
Listed below are some typical chemical formulas:
- H₂O, or water
- NaCl: Salt
Baking soda is NaHCO₃.
- NaClO for bleach
- Table sugar (sucrose): C12H22O11
- CO₂: Carbon dioxide
- NH₃ for ammonia
- H₂O₂, or hydrogen peroxide
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a student spotted a tlc plate and ran it in 10% ethyl acetate/hexanes. the tlc obtained showed a streak rather than separate spots for the components. what technical mistake might the student have made?
To fix the issue, the student should carefully reapply the sample in a smaller amount, allow it to dry completely, and adjust the solvent system if necessary to achieve proper separation of the components on the TLC plate.
The student might have made the following technical mistake while running the TLC plate:
1. Overloading the sample: When spotting the TLC plate, the student may have applied too much sample, causing the components to streak rather than separate into individual spots. To resolve this, the student should apply a smaller amount of sample and ensure that it is evenly distributed.
2. Insufficient drying: If the student did not allow the spotted sample to dry properly before placing it in the solvent, it can cause the components to streak. To prevent this, the student should ensure the sample is completely dry before running the TLC plate.
3. Inappropriate solvent system: Although the student used a 10% ethyl acetate/hexane mixture, it is possible that this solvent system was not suitable for the specific sample. Adjusting the solvent ratio or trying different solvent systems could help achieve better separation.
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Elemental analysis of a pure compound indicated that the compound contained 324 g of C, 48.5 g of H and 16.0 g of O. What is its empirical formula?
Answer:
To find the empirical formula of a compound, we need to determine the simplest whole number ratio of the atoms present in the compound. We can do this by dividing each element's mass by its molar mass to get the number of moles of each element, and then dividing each number of moles by the smallest number of moles obtained. The molar masses of carbon, hydrogen, and oxygen are 12.01 g/mol, 1.008 g/mol, and 16.00 g/mol, respectively. Number of moles of C = 324 g / 12.01 g/mol = 26.98 mol Number of moles of H = 48.5 g / 1.008 g/mol = 48.11 mol Number of moles of O = 16.0 g / 16.00 g/mol = 1.
Answer:
C27H48O
Explanation:
To determine the empirical formula of the compound, we need to find the simplest whole number ratio of atoms in the compound. We can do this by assuming that we have 100 g of the compound, and finding the number of moles of each element in this amount.
Number of moles of carbon (C): 324 g / 12.01 g/mol = 26.98 mol
Number of moles of hydrogen (H): 48.5 g / 1.01 g/mol = 48.02 mol
Number of moles of oxygen (O): 16.0 g / 16.00 g/mol = 1.00 mol
Next, we divide each of these mole values by the smallest value to get the simplest ratio:
C: 26.98 mol / 1.00 mol = 26.98
H: 48.02 mol / 1.00 mol = 48.02
O: 1.00 mol / 1.00 mol = 1.00
We can see that the simplest ratio of atoms in the compound is approximately C27H48O. However, we need to express this as a whole number ratio, so we divide each subscript by the smallest subscript (which is 1):
Empirical formula: C27H48O
Therefore, the empirical formula of the compound is C27H48O.
How can you differentiate Hydrogen and carbon dioxide gases. (by flame)
The presence of hydrogen would be indicated by a pale blue flame that is nearly invisible in broad daylight, but the presence of carbon dioxide would be indicated by the flame going out.
Why do carbon dioxide and hydrogen gas flames differ from one another?A flame ignited by hydrogen gas emits a barely perceptible pale blue flame under normal lighting conditions. This is due to the flame that hydrogen gas produces mostly emitting light in the ultraviolet spectrum, which is invisible to the human eye.
On the other side, a flame is put out when carbon dioxide gas is added to it. This is because carbon dioxide is a gas that does not support burning and is not flammable.
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write the net ionic equation for the acid-base hydrolysis equilibrium that is established when ammonium nitrate is dissolved in water. (use h3o instead of h .)
The net ionic equation for the acid-base hydrolysis equilibrium that is established when ammonium nitrate is dissolved in water is given below:Answer: NH4+(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq) + NO3-(aq).
The ionic equation is a chemical equation in which the electrolytes in aqueous solution are represented by their actual ions rather than their complete formulas. It indicates that ions undergo a chemical reaction to produce a new compound.
The net ionic equation displays the actual chemical reaction taking place in an aqueous solution. It is derived by eliminating spectator ions, which do not play any active role in the chemical reaction.The acid-base hydrolysis equilibrium is as follows:NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq).
The ammonium ion hydrolyzes to form ammonium ion and hydronium ions. The nitrate ion is a spectator ion that does not participate in the reaction. Therefore, the net ionic equation is given by:NH4+(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq).
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what is the concentration of hcl after diluting 10 ml of concentrated hcl solution (25% with a density of 1.07 g/ml) into a 250 ml volumetric flask? e) what is the concentration of nh3 after diluting 15 ml of concentrated nh3 (15% with a density of 0.75 g/ml) into a 100 ml volumetric flask?
HCl after dilution: Final concentration = 0.107 g/ml, NH₃ after dilution: Final concentration = 0.016875 g/ml
To calculate the concentration of HCl after dilution:
Step 1: Find the mass of HCl in the 10 ml concentrated solution.
Mass = Volume × Density × Concentration
Mass = 10 ml × 1.07 g/ml × 0.25
Mass = 26.75 g
Step 2: Calculate the final volume after dilution, which is the volume of the volumetric flask.
Final volume = 250 ml
Step 3: Calculate the final concentration of HCl.
Final concentration = Mass / Final volume
Final concentration = 26.75 g / 250 ml
Final concentration = 0.107 g/ml (or 10.7% by mass)
To calculate the concentration of NH₃ after dilution:
Step 1: Find the mass of NH3 in the 15 ml concentrated solution.
Mass = Volume × Density × Concentration
Mass = 15 ml × 0.75 g/ml × 0.15
Mass = 1.6875 g
Step 2: Calculate the final volume after dilution, which is the volume of the volumetric flask.
Final volume = 100 ml
Step 3: Calculate the final concentration of NH₃.
Final concentration = Mass / Final volume
Final concentration = 1.6875 g / 100 ml
Final concentration = 0.016875 g/ml (or 1.6875% by mass)
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the name given to an aqueous solution of hbr is . group of answer choices bromic acid hypobromous acid hydrogen bromide bromous acid hydrobromic acid
The name given to an aqueous solution of HBr is "hydrobromic acid." Option E is correct.
Hydrobromic acid (HBr) is a strong acid that forms when hydrogen bromide gas dissolves in water. It is a clear, colorless liquid having a pungent odor as well as it is highly corrosive. Hydrobromic acid is commonly used in the laboratory and in various industrial applications, including the production of pharmaceuticals, dyes, and other chemicals. It is also used as a reagent in organic chemistry for various types of reactions.
An aqueous solution is the solution in which water will be the solvent. In other words, it is a solution where water is the substance that dissolves other substances, which are called solutes. Many substances can dissolve in water to form aqueous solutions, including salts, acids, and bases.
Hence, E. is the correct option.
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--The given question is incomplete, the complete question is
"The name given to an aqueous solution of HBr is . group of answer choices A) bromic acid B) hypobromous acid C) hydrogen bromide D) bromous acid E) hydrobromic acid."--
calculate the volume of 5.9 x 10^23 molecules of propane gas trapped in a container at a pressure of 253.3 kpa and a temp
To calculate the volume of
[tex]5.9 \times 10^23[/tex]
molecules of propane gas trapped in a container at a pressure of 253.3 kPa and a temperature, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.
We need to convert the number of molecules to moles. The molar mass of propane is 44.1 g/mol, so the number of moles of propane is
[tex]5.9 \times 10^23[/tex]
molecules /
[tex]6.022 \times 10^23[/tex]
molecules/mol = 0.98 moles.
We need to convert the pressure to atmospheres (atm), which is the unit typically used with the ideal gas law. 253.3 kPa / 101.3 kPa/atm = 2.50 atm.
We also need to convert the temperature to Kelvin by adding 273.15 to the Celsius temperature. Let's assume the temperature is 25°C, so T = 25°C + 273.15 = 298.15 K.
We can plug in the values into the ideal gas law equation and solve for V:
[tex]V = (nRT) / P = (0.98 mol \times 0.0821 L•atm/mol•K \times 298.15 K) / 2.50 atm = 29.6 L[/tex]
The volume of
[tex]5.9 \times 10^23[/tex]
molecules of propane gas trapped in a container at a pressure of 253.3 kPa and a temperature of 25°C is 29.6 L.
The ideal gas law equation is a useful tool to calculate the volume of a gas sample when its pressure, temperature, and amount of substance are known.
It is important to convert the units to the appropriate ones and use the correct value for the gas constant depending on the units used.
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To determine the temperature water becomes ice, Passaic DPW employees
made a salt water solution. If you made a solution containing 0.50 mole of
rock salt in 300 g of water. What is the new freezing point of the solution?
the new freezing point of the solution is 0°C - 6.19°C = -6.19°C.
Assuming complete dissociation of the rock salt in water, the number of particles in the solution is 0.50 moles of rock salt x 2 particles per formula unit (NaCl) = 1.0 mole of particles.
The molal concentration of the solution can be calculated as follows:
molality (m) = moles of solute / mass of solvent in kg
mass of solvent = 300 g = 0.3 kg
m = 1.0 moles / 0.3 kg = 3.33 m
The freezing point depression (ΔTf) of the solution can be calculated using the formula:
ΔTf = Kf x m
where Kf is the freezing point depression constant of water, which is 1.86 °C/m.
ΔTf = 1.86 °C/m x 3.33 m = 6.19 °C
Therefore, the new freezing point of the solution is 0°C - 6.19°C = -6.19°C.
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Help what's the answer??
Answer: 1.57g and 70.5%
Explanation:
Theoretical Yield
The theoretical yield of a reaction is the absolute maximum amount of product that could be created with the amounts of reactants.
This problem gives us the amount of hydrochloric acid, which is 6.37 grams. The molar mass of HCl is the molar mass of hydrogen plus the molar mass of chlorine, which is 36.46 g/mol.
To find the moles of HCl, we just divide the mass by the molar mass.
6.37/36.46 = 0.175 moles HCl
Since oxygen is in excess, the amount used in the reaction will be dictated by the amount of HCl used in the reaction. It does not need to be taken into consideration when determining the amount of reactant since it is in excess.
To find the theoretical yield of water, we will do stoichiometry.
Balancing the equation
Written out with the chemical symbols, this equation is
HCl + O2 ⇒H2O + Cl
This is not balanced, since there is 1 hydrogen on the left side and 2 on the right, and 2 oxygens on the left and 1 on the right.
To balance this, we can put coefficients in front of some reactants and products to make sure there are equal amounts of everything on each side.
The balanced equation will be 4HCl + O2 ⇒ 2H2O + Cl
Now, there are 4 hydrogens on the left and 4 on the right, as well as 2 oxygens on the left and 2 on the right. It is balanced.
We can see by looking at the coefficients of the balanced equation that every 4 moles of HCl consumed will produce 2 mole of H2O, so the ratio is 1:2.
To do stoichiometry, we will multiply the moles of HCl by the ratio of H2O to HCl, which is just dividing by 2.
The theoretical yield of water is then 0.175 moles HCl * [tex]\frac{1moleH2O}{2moleHCl}[/tex] = 0.1874 moles H2O.
Our theoretical yield is 0.0874 moles H2O. But the question and the actual yield are in grams, so we will convert this to grams. To convert moles to grams, just multiply the moles by the molar mass. The molar mass of water is 18.0 g/mol, so
0.0874*18.0 = 1.57 g
The theoretical yield is 1.57 g H2O
Percent Yield
Percent yield is much easier. Percent yield is
((actual yield)/(theoretical yield))*100
In this case, our actual yield is 1.11 grams and our theoretical yield is 3.15 grams, so
[tex]\frac{1.11}{1.57} =[/tex] 70.5%
70.5% is the percent yield.
1) How many moles of gas occupy 58 L at a pressure of 1.55 atmospheres and a temperature of 222 K?
To find the moles of the gas , we can use the ideal gas law. Which states -
[tex] \:\:\:\:\:\:\:\:\:\star\longrightarrow \sf \underline{PV=nRT} \\[/tex]
Where:-
P is the pressure measured in atmospheres V is the volume measured in litersn is the number of moles.R is the ideal gas constant (0.0821 L atm mol⁻¹ K⁻¹).T is the temperature measured in kelvin.As per question, we are given that-
P=1.55 atmV= 58 LT = 222 KR = 0.08206 L atm mol⁻¹ K⁻¹Now that we have all the required values, so we can put them all in the Ideal gas law formula and solve for moles -
[tex] \:\:\:\:\:\:\:\:\:\star\longrightarrow \sf \underline{PV=nRT} \\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf 1.55 \times 58 = n \times 0.0821 \times 222\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf 89.9 = n \times 18.2262\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\longrightarrow \sf n \times 18.2262 =89.9\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf n = \dfrac{89.9}{18.2262}\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\:\:\:\longrightarrow \sf n =4.9324......\\[/tex]
[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf \underline{n =4.93 \:moles }\\[/tex]
Therefore, 4.93 moles of gas will be occupied 58 L at a pressure of 1.55 atmospheres and a temperature of 222K.
Answer:
4.93 moles
Explanation:
To find how many moles of gas pressure occupy 58 L at a pressure of 1.55 atmospheres and a temperature of 222 K, use the ideal gas law.
Ideal Gas Law[tex]\boxed{\sf PV=nRT}[/tex]
where:
P is the pressure measured in atmospheres (atm).V is the volume measured in liters (L).n is the number of moles.R is the ideal gas constant (0.08206 L atm mol⁻¹ K⁻¹).T is the temperature measured in kelvin (K).As we are solving for the number of moles, rearrange the equation to isolate n:
[tex]\implies \sf n=\dfrac{PV}{RT}[/tex]
Given values:
P = 1.55 atmV = 58 LR = 0.08206 L atm mol⁻¹ K⁻¹T = 222 KSubstitute the values into the formula and solve for n:
[tex]\implies \sf n=\dfrac{1.55 \cdot 58}{0.08206 \cdot 222}[/tex]
[tex]\implies \sf n=\dfrac{89.9}{18.21732}[/tex]
[tex]\implies \sf n=4.93\;mol\; (3\;s.f.)[/tex]
Therefore, 4.93 moles of gas occupy a volume of 58 L at a pressure of 1.55 atm and a temperature of 222 K.
I have 0.60 moles of sodium iodide (NaI). How many liters of water would it take to make a 0.19 M solution?
it would take 3.16 litres of water to make a 0.19 M solution of NaI from 0.60 moles of NaI.
To calculate the volume of water needed to make a 0.19 M solution of NaI, we need to use the formula:
Moles of solute = Molarity x Volume (in liters)
We can rearrange this formula to solve for the volume of water:
Volume (in liters) = Moles of solute / Molarity
First, let's calculate the number of moles of NaI in 0.60 moles:
Moles of NaI = 0.60 moles
Now, we can use the formula above to calculate the volume of water needed:
Volume (in litres) = Moles of NaI / Molarity
Volume (in litres) = 0.60 moles / 0.19 M
Volume (in litres) = 3.16 litres
Therefore, it would take 3.16 litres of water to make a 0.19 M solution of NaI from 0.60 moles of NaI.
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Given the following equation:
2H2O --> 2H2 +O2
What mass of oxygen would form from 5 moles of water?
Question 6 options:
.078
320
.3125
80
A student dilutes 15.00 mL of 0.275 M NaNO3 stock solution to a volume of 100.0 mL. What is the final molarity?
When a stock solution is diluted, the number of moles of solute (NaNO3 in this case) remains constant. Therefore, we can use the following equation to find the final molarity of the diluted solution:
M1V1 = M2V2
where M1 is the initial molarity (0.275 M), V1 is the initial volume (15.00 mL), M2 is the final molarity (what we want to find), and V2 is the final volume (100.0 mL).
First, we need to convert the initial volume from milliliters to liters:
V1 = 15.00 mL = 0.01500 L
Next, we can substitute the given values into the equation:
(0.275 M) × (0.01500 L) = M2 × (0.1000 L)
Solving for M2, we get:
M2 = (0.275 M × 0.01500 L) ÷ 0.1000 L
M2 = 0.04125 M
Therefore, the final molarity of the NaNO3 solution is 0.04125 M.
Need help please!! Don’t know any of this.
1. The mass (in grams) of FeBr₃ produced from 65 g of Br₂ is 80.13 g
2. The mole of CO₂ formed from 10 moles of C₅H₁₂ is 50 moles
3. The moles of MnCl₂ prepared from 52.1 grams of MnO₂ is 0.6 mole
1. How do i determine the mass of FeBr₃ produced?The mass of FeBr₃ produced can be obtained as illustrated below:
2Fe + 3Br₂ -> 2FeBr₃
Molar mass of Br₂ = 160 g/molMass of Br₂ from the balanced equation = 3 × 160 = 480 g Molar mass of FeBr₃ = 295.85 g/molMass of FeBr₃ from the balanced equation = 2 × 295.85 = 591.7 gFrom the balanced equation above,
480 g of Br₂ reacted to produce 591.7 g of FeBr₃
Therefore,
65 g of Br₂ will react to produce = (65 × 591.7) / 480 = 80.13 g of FeBr₃
Thus, the mass of FeBr₃ produced is 80.13 g
2. How do i determine the mole of CO₂ formed?The mole of CO₂ formed can be obtained as follow:
C₅H₁₂ + 8O₂ -> 5CO₂ + 6H₂O
From the balanced equation above,
1 mole of C₅H₁₂ reacted to produce 5 moles of CO₂
Therefore,
10 moles of C₅H₁₂ will react to produce = 10 × 5 = 50 moles of CO₂
Thus, the mole of CO₂ formed is 50 moles
3. How do i determine the mole of MnCl₂ prepared?First, we shall obtain the mole of 52.1 g of MnO₂, Details below:
Mass of MnO₂ = 52.1 grams Molar mass of MnO₂ = 86.94 g/mol Mole of MnO₂ =?Mole = mass / molar mass
Mole of MnO₂ = 52.1 / 86.94
Mole of MnO₂ = 0.6 mole
Finally, we shall determine the mole of MnCl₂ prepared can be obtained as follow:
MnO₂ + 4HCl -> MnCl₂ + Cl₂ + 2H₂O
From the balanced equation above,
1 mole of MnO₂ reacted to produce 1 mole of MnCl₂
Therefore,
0.6 mole of MnO₂ will also react to produce 0.6 mole of MnCl₂
Thus, the mole of MnCl₂ prepared is 0.6 mole
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which interactions can contribute to the intrinsic binding energy during enzymatic catalysis?electrostatic interactionspermanent covalent bondingvan der waals interactionsnucleophilic attack by serinehydrogen bonding
Intrinsic binding energy during enzymatic catalysis is caused by a variety of interactions.
Here are the interactions that can contribute to the intrinsic binding energy during enzymatic catalysis:
Electrostatic interactions are caused by the attraction of opposite charges or the repulsion of like charges. Enzymatic catalysis can be influenced by these interactions.
Permanent covalent bonding is a type of bonding that involves the sharing of electrons between two atoms. The formation of a covalent bond can help in the catalytic process.
Van der Waals interactions are a type of intermolecular force that arises due to fluctuations in the electron density around an atom. These interactions can also contribute to the intrinsic binding energy during enzymatic catalysis.
Nucleophilic attack by serine is a reaction that is commonly used in enzymatic catalysis. The serine acts as a nucleophile and attacks the substrate molecule, which results in the formation of a covalent bond between the enzyme and the substrate molecule.
Hydrogen bonding is another type of interaction that can contribute to the intrinsic binding energy during enzymatic catalysis. Hydrogen bonds are formed between the enzyme and the substrate molecule, which can help to stabilize the transition state during the catalytic reaction.These are the interactions that can contribute to the intrinsic binding energy during enzymatic catalysis.
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The diagram shows part of a DNA molecule. Using the order of bases in the top strand, write the letters of the bases that belong on the bottom strand.
Answer:
G, A, A, T, C, C, G, A, A, T, G, G, T
Explanation:
What is the molar mass of an unknown if a 0.45 M solution is created by dissolving 12 grams in 425 mL of water?
To calculate the molar mass of the unknown substance, we need to use the formula:
Molar mass = (mass of solute) / (number of moles of solute)
First, let's calculate the number of moles of solute in the solution:
Number of moles = (concentration) x (volume in liters)
We know that the concentration of the solution is 0.45 M, and the volume of the solution is 425 mL, which is equivalent to 0.425 L. Substituting these values into the formula, we get:
Number of moles = 0.45 M x 0.425 L
Number of moles = 0.19125 moles
Next, we can calculate the mass of the solute (the unknown substance) by using the formula:
mass = number of moles x molar mass
Rearranging the formula, we get:
molar mass = mass / number of moles
We know that the mass of the solute is 12 grams, and we have already calculated the number of moles as 0.19125 moles. Substituting these values into the formula, we get:
molar mass = 12 g / 0.19125 moles
molar mass = 62.8 g/mol
Therefore, the molar mass of the unknown substance is 62.8 g/mol.
if i add 25 ml of ater to 125 ml of a 0.15 m sodium hydroxide solution, what will the molarity of the diluted solution be?
The molarity of the diluted solution is 0.125M.
To find the molarity of the diluted solution after adding 25 ml of water to 125 ml of a 0.15 M sodium hydroxide solution, you can follow these steps:
1. Determine the initial volume (V1) and molarity (M1) of the sodium hydroxide solution: V1 = 125 ml and M1 = 0.15 M.
2. Determine the volume of water added (V2): V2 = 25 ml.
3. Calculate the total volume of the diluted solution (Vt):
Vt = V1 + V2
Vt = 125 ml + 25 ml
Vt = 150 ml.
4. Use the dilution equation M1V1 = M2V2, where M2 is the molarity of the diluted solution.
5. Solve for M2:
M2 = (M1V1) / Vt
M2 = (0.15 M × 125 ml) / 150 ml
M2 = 18.75 / 150
M2 = 0.125 M.
So, the molarity of the diluted solution after adding 25 ml of water to 125 ml of a 0.15 M sodium hydroxide solution will be 0.125 M.
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5. How many atoms are found in a 15.5 g sample of bismuth (Bi)? See periodic
table for the molar mass of bismuth. Use dimensional analysis, show all work to
receive full credit. (5 pts)
a. 9.33 × 1024 atoms
b.
3.24 × 10³ atoms
1.26 x 1022 atoms
d. 4.46 × 1022 atoms
C.
which of the following correctly identifies the dependent and independent variables in this experiment? responses the color of the light is the dependent variable, and the percentage of plants showing phototropism is the independent variable. the color of the light is the dependent variable, and the percentage of plants showing phototropism is the independent variable. the percentage of plants showing phototropism is the dependent variable, and the color of the light is the independent variable. the percentage of plants showing phototropism is the dependent variable, and the color of the light is the independent variable. the direction of the light is the dependent variable, and the percentage of plants showing phototropism is the independent variable. the direction of the light is the dependent variable, and the percentage of plants showing phototropism is the independent variable. the color of the light is the dependent variable, and the direction of the light is the independent variable.
The percentage of plants showing phototropism is the dependent variable, and the color of the light is the independent variable. Therefore, the correct answer is: "the percentage of plants showing phototropism is the dependent variable, and the color of the light is the independent variable."
The dependent variable is the variable that is being measured or observed, and its value depends on the independent variable, which is the variable that is being manipulated or changed in the experiment. In this experiment, the percentage of plants showing phototropism is being measured, which means that it is the dependent variable. The color of the light is being manipulated, which means that it is the independent variable.
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the ph of a solution of hexanoic acid is measured to be . calculate the acid dissociation constant of hexanoic acid. be sure your answer has the correct number of significant digits.
The acid dissociation constant of hexanoic acid is 4.93 × 10^-10 mol/L.
When measuring the pH of a solution of hexanoic acid to be 4.96, the acid dissociation constant of hexanoic acid can be calculated. This can be done through the following equation:
Ka = [H3O+][A-] / [HA]whereKa
= acid dissociation constantH3O+
= hydronium ionA-
= conjugate baseHA
= acidThe pH of the solution of hexanoic acid is measured to be 4.96.
Thus, [H3O+] is equal to 10^-4.96 or 7.02 × 10^-5 M.
The initial concentration of the hexanoic acid is equal to the concentration of the undissociated acid or [HA].The acid dissociation constant of hexanoic acid can be calculated by plugging the known values into the equation:
Ka = [7.02 × 10^-5][A-] / [HA]The concentration of the conjugate base, A-, is equal to the concentration of the dissociated acid, which is equal to [H3O+].
Thus,Ka = [7.02 × 10^-5]^2 / [HA]Ka = 4.93 × 10^-10 mol/L
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