The balanced chemical equation for the reaction is 4Na + O₂ → 2Na₂O. This means that for every 4 moles of sodium (Na) and 1 mole of oxygen (O₂) that react, 2 moles of sodium oxide (Na₂O) are produced.
To determine the correct answer, we need to use stoichiometry to calculate the theoretical yield of Na₂O based on the amount of Na and O₂ present in each scenario. We can then compare the calculated yield to the given yield of 169 g and 527 g to see which starting material (Na or O₂) produces a lower yield.
A. To calculate the theoretical yield of Na₂O based on 5.43 moles of Na:
5.43 moles Na x (2 moles Na₂O / 4 moles Na) x (62 g Na₂O / 1 mole Na₂O) = 168.78 g Na₂O
The calculated yield of Na₂O based on 5.43 moles of Na is very close to the given yield of 169 g. Therefore, we can conclude that Na is not the correct answer.
B. To calculate the theoretical yield of Na₂O based on 4.25 moles of O₂:
4.25 moles O₂ x (2 moles Na₂O / 1 mole O₂) x (62 g Na₂O / 1 mole Na₂O) = 527.25 g Na₂O
The calculated yield of Na₂O based on 4.25 moles of O₂ is very close to the given yield of 527 g. Therefore, we can conclude that O₂ is not the correct answer.
C. Since Na is not the correct answer and O₂ is not the correct answer, the only option left is C. Therefore, the correct answer is O₂ because it has the lower starting mass.
the radioactive substance iridium-192 has a half-life of 73.83 days. how many days would it take for a sample of iridium-192 to decay to 10% of its original amount? round your answer to the nearest integer.
It would take 221 days for a sample of iridium-192 to decay to 10% of its original amount.
The half-life of iridium-192 is 73.83 days.
Now, the formula for calculating the time is given by;
ln(Nt/N0)=-λt
Where Nt is the number of radioactive atoms remaining after time t, N0 is the initial number of radioactive atoms, t is the half-life and λ is the decay constant.
Taking 10% of the initial amount of the radioactive substance which is the final amount of the substance and solving for t gives
ln(0.1) = -λt
Now substituting the value of half-life which is 73.83 days into the equation gives;
ln(0.1) = -ln(2)/73.83(t)= -ln(0.1) / ln(2) * 73.83t = 220.64 days (rounded to the nearest integer).
Thus, it would take 221 days for a sample of iridium-192 to decay to 10% of its original amount.
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Sodium carbonate and calcium chloride are mixed, which of the following
correctly describes the outcome?
a. No reaction is seen
b. A precipitate of calcium carbonate forms
c. A precipitate of sodium chloride forms
d. A precipitate of sodium and calcium is seen.
When sodium carbonate and calcium chloride are mixed, a precipitate of calcium carbonate forms.
Option B is correct.
What is a chemical reaction?
A chemical reaction is described as a process that leads to the chemical transformation of one set of chemical substances to another.
The reaction between sodium carbonate (Na2CO3) and calcium chloride (CaCl2) is a double displacement reaction, which is shown by the following chemical equation:
Na2CO3 + CaCl2 → CaCO3 + 2NaCl
We can see that the carbonate ion (CO32-) from sodium carbonate combines with the calcium ion (Ca2+) from calcium chloride to form solid calcium carbonate (CaCO3), which is insoluble in water and appears as a precipitate.
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please help me balance the following equations
which metal(s) can be oxidized with a sn2 solution but not with an fe2 solution? check all that apply.
The metals which can be oxidized with a Sn²⁺ solution but not with a Fe²⁺ solution are nickel, cadmium and copper.
The reactivity order of the metals is given below:
magnesium, aluminium, zinc, iron, nickel and tin
Generally, oxidation is defined as the loss of electrons and more reactive metals gives away electrons to less reactive metals.
So we are looking for a metal higher than tin but lower than iron.
The only metal in the rhyme is nickel. Definitely not aluminium because that is higher than iron.
Hence, the answer is Nickel and cadmium. Also copper (Cu) is one such metal, which has a reduction potential of -0.34 V. Hence, Cu can be oxidized by Sn2+ but not by Fe2+.
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What types of changes do you observe between the rock formations in the two images?
In general, changes in rock formations can occur due to a variety of factors, including weathering, erosion, tectonic activity, and sedimentation.
What is Rock Formation?
Rock formation refers to the process by which rocks are created, transformed, or modified over time through various geological processes. This process involves the deposition of sediments, the consolidation and hardening of sediments into rock layers, and the metamorphism of existing rocks due to heat, pressure, and other factors.
There are three main types of rocks: igneous, sedimentary, and metamorphic. Igneous rocks form from molten magma or lava that cools and solidifies, while sedimentary rocks are formed from the accumulation of sediments, such as sand, mud, and organic material. Metamorphic rocks are formed from the transformation of existing rocks under high heat, pressure, and chemical reactions.
Weathering and erosion can cause rocks to break down and change shape over time, while tectonic activity can cause rocks to shift and deform. Sedimentation can lead to the formation of new rock layers on top of existing ones.
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What is the ph of a solution of 0.20 m hno2 containing 0.10 m nano2 at 25°c, given k a of hno2 is 4.5 × 10–4?
The pH of the given solution is 2.74 at 25°C. The pH of a solution of 0.20 M HNO2 containing 0.10 M NaNO2 at 25°C can be calculated using the Ka value of HNO2. HNO2 is a weak acid and dissociates in water to form H+ and NO2-.
The Ka expression for this reaction is Ka = [H+][NO2-]/[HNO2]. Since the concentration of NaNO2 is much larger than that of HNO2, we can assume that the concentration of HNO2 does not change significantly due to the dissociation. Therefore, we can use the initial concentration of HNO2 in the Ka expression. Substituting the given values into the expression and solving for [H+], we get [H+] = 1.8 × 10^-3 M. Taking the negative logarithm of this value gives the pH of the solution, which is approximately 2.74.
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Calculate the pH of a soultionprepared by dissolving 0.750 mol of NH3 and 0.250 mol of NH4Cl in water sufficnet yield 1.00L of solution. The Kb of ammonia is 1.77 x 10^-5
According to the given statement The pH of the solution is 9.72.
Where can you find ammonia?Ammonia (NH3) is a substance that may be found in the air, soil, water, as well as in plants, animals, and people. Several commercial and domestic cleansers include ammonia as well. Ammonia at high concentrations can irritate and burn the eyes, mouth, throat, lungs, and skin.
This query may be addressed using the Henderson-Hasselbalch formula:
pH = pka + log [Base] /[Acid]
Base is NH₃ and acid NH₄⁺
Molarity of the compounds is:
NH₃: 0.750mol / 1.00L = 0.750M
NH₄⁺: 0.250mol / 1.00L = 0.250M
To find pka:
Ka×Kb = Kw
Ka = 1x10⁻¹⁴ / 1.77x10⁻⁵ = 5.65x10⁻¹⁰
pKa = -logKa = 9.25
Replacing:
pH = 9.25 + log [0.750] /[0.250]
pH = 9.72
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calculate the ph during the titration of 22.84 ml of 0.26 m hno3(aq) with 0.10 m naoh after 11.72 ml of the base have been added.
explain why k values for these four reactions should all be approximately equal?
The value of k i.e. rate constant indicate the reaction is very fast. In this case, it is the first-order reaction.
The higher value in this case is the order of 12. The unit of k is (1/time) i.e. (1/s). The rate expression is given as follows
[tex]r_{A}=\frac{dC_{A}}{dt}=-kC_{A}..............(1)dC = -kdt..............................(2)[/tex]
If we consider the case of CA, for example, either Iodide ions as their concentration are small as compared to BrO3- and hydrogen ions, then iodide ion concentration is decreasing very fast.
Similar is the case for others. So, it increases the rate of reaction of the order of 12. The rate of disappearance of reactants mentioned in the table is almost high and it shows nearly the same value of k.
2. The rate of reaction is based on the concentration of the limiting reactant.
The rate of reaction is given by the following expression. Integrating equation(2) from initial concentration (CA0) to final concentration (CA),
[tex]\int_{C_{A0}}^{C_{A}}-\frac{dC_{A}}{C_{A}} = k\int_{0}^{t}dt-ln\frac{C_{A}}{C_{A0}}=kt[/tex]
There is no information about the initial concentration of reactants (CA0) i.e. initial volume of stock solution for each of the reactants.
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The complete question is:
Rates of Chemical Reactions: A Clock Reaction tab/-R e p o r t Name Time MTWRF Reaction # k" k, 3.68 x 10,2 Explain why values for these four reactions should all be approximately equal: Use the k above and the reactant concentrations from Part A to predict the relative rate and time (t) for reaction mixture #5 (show work): trek" relative rateprekted
needing help with chemistry !
Name the three items that will change the solubility of a solute into a solvent
Name three way to increase the speed with which a solute dissolves.
Three items that will change the solubility of a solute into a solvent are:
1. Temperature - increasing temperature typically increases the solubility of solids in liquids, but can decrease the solubility of gases in liquids.
2. Pressure - increasing pressure can increase the solubility of gases in liquids.
3. Polarity - solutes that have similar polarity to the solvent are more likely to dissolve.
Three ways to increase the speed with which a solute dissolves are:
1. Stirring or agitating the solution to increase the surface area of the solute in contact with the solvent.
2. Increasing the temperature of the solvent, which can increase the kinetic energy of the solvent molecules and the solute particles, leading to more frequent collisions and faster dissolution.
3. Grinding or crushing the solute to decrease the particle size and increase the surface area in contact with the solvent.
LESSON 1
Content Practice B
1
Position and Motion
Directions Complete these purphs by writing the correct terms on the lines. Some terms might be used more
you must first choose ain)
To describe an object's (1. )
(2. )
as a starting place. From there, you must specify the
in
(3. )
to the object and the (4. )
which it lies from the starting place. If you are giving directions to two objects located
it can sometimes
in different directions from the same (5. )
direction
be helpful to describe one object as being in the (6. )
direction
from that place and the other in the (7. )
An object is in (8. )
any time its
is changing. In most cases, such a change involves changes in
(10. )
and (11. )
from the starting
point. However, if an object returns to its starting point, its
(12. )
is zero, even though it might have traveled
The passage describes how to describe an object's position and motion, including specifying its direction and distance from a starting point. It also notes that an object is in motion when its position is changing, and that an object's displacement can be zero even if it has traveled.
How to describe an object's location and motion is covered in the paragraph. It states that it is crucial to indicate an object's direction and distance from a starting point when expressing an object's location. To achieve this, one may define the direction of the item using words like north, south, east, and west, and its distance from the beginning point using words like metres or feet. The verse also indicates that when an object's location changes, it is in motion. Both the direction and the distance from the starting location may alter as a result. Even when an item has moved, its displacement might still be zero.
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Identify the compound with atoms that have an incomplete octet.
A) BF3
B) ICl5
C) CO2
D) CO
E) Cl2
The compound with atoms that have an incomplete octet is BF3.The correct answer is a.
An octet is a set of eight electrons in the outermost shell of an atom. Noble gases have an octet of electrons in their outermost shells, making them stable. Other elements aim to reach this stable state by either losing or gaining electrons. When they do this, they form ions.
In some cases, however, elements may share electrons to achieve an octet. This is called covalent bonding. In a covalent bond, atoms share valence electrons to reach the stable octet configuration.BF3 is an example of a compound that has atoms with an incomplete octet.
In BF3, boron has only six electrons in its outermost shell. This means that it cannot form an octet on its own. However, by sharing three electrons with three fluorine atoms, boron is able to achieve a stable configuration, even if it does not have a complete octet.
In contrast, the other compounds listed in the question all have atoms that have a complete octet or an expanded octet. For example, CO and CO2 both have atoms with a complete octet. ICl5 and Cl2 both have atoms with an expanded octet. Only BF3 has atoms with an incomplete octet.
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the concentration of barium ion in a solution is 0.010 m. what concentration of sulfate ion is required to just begin precipitating baso4?
The concentration of sulfate ion required to begin precipitation of BaSO₄ is [tex]1.1 x 10^{-8} M[/tex]. Any concentration of sulfate ion greater than this will result in precipitation of BaSO₄.
The solubility product constant (Ksp) is an equilibrium constant that represents the concentration of the ions in a saturated solution at a given temperature. When the concentration of any ion exceeds the Ksp, the excess ions begin to form a solid, and precipitation occurs. For BaSO₄, the Ksp value is [tex]1.1 x 10^{-10}[/tex] at 25°C.
To calculate the concentration of sulfate ion required to begin precipitation of BaSO₄, we can use the Ksp expression and the initial concentration of barium ion:
BaSO₄(s) ↔ Ba₂+(aq) + SO₄₂–(aq)Ksp = [Ba₂+][SO₄₂–]Substituting the given concentration of barium ion (0.010 M) and the Ksp value, we get:
[tex]1.1 x 10^{-10}[/tex]= (0.010 M)[SO₄₂–][SO₄₂–] = [tex]1.1 x 10^{-8}[/tex] MTherefore, the concentration of sulfate ion required to begin precipitation of BaSO₄ is [tex]1.1 x 10^{-8}[/tex] M. Any concentration of sulfate ion greater than this will result in precipitation of BaSO₂.
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does the rate constant for a reaction generally increase or decrease with an increase in reaction temperature? which factor is most sensitive to changes in temperature? complete the sentences to explain your answer.
The rate constant for a reaction generally increases with an increase in reaction temperature. The factor most sensitive to changes in temperature is the activation energy of the reaction.
The activation energy is the minimum energy required for a reaction to occur.
According to the Arrhenius equation, the rate constant (k) is directly proportional to the temperature (T) of the system. An increase in temperature would lead to an increase in the kinetic energy of the reactant molecules which will result in an increase in the number of effective collisions.
These effective collisions will have energy greater than the activation energy barrier, and thus they will lead to the formation of products. Hence, the rate of the reaction will increase with an increase in temperature.
Most of the reactions have a positive value of activation energy. Therefore, a slight increase in temperature can increase the rate of the reaction. However, if the temperature is increased beyond a certain limit, the rate of the reaction decreases due to the thermal denaturation of the enzyme or the inactivation of the catalysts, or the destruction of the reactants.
On increasing or decreasing the temperature the activation energy of a reaction changes drastically, either decreasing or increasing respectively.
In summary, the rate constant for a reaction is directly proportional to temperature and the activation energy is most sensitive toward temperature.
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What type of energy is stored in the nucleus of a uranium atom?
when sugar is mixed with water what does it form?
The answer is Solution!
Explanation:
Solution is a homogeneous mixture
of two or more substance. In a
sugar solution, sugar gets uniformly
mixed with water
1.275 X 10^25 molecules of O₂ to grams
From the calculations, we can see that the number of moles of the oxygen molecules that is present is 21.2 moles.
What is the number of moles?The moles refers to the n umber of the elementary entities that we have in the substances and in this case we are dealing with the number of moles that we have in the oxygen molecule here.
We know that;
1 mole of O2 contains 6.02 * 10^23 molecules
x moles of the O2 contains 1.275 X 10^25 molecules of O₂
x = 21.2 moles
Thus what we have is about 21.2 moles of oxygen
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in each of the following reactions, the aromatic ring has just one chemically distinct, aromatic h, so a single electrophilic aromatic substitution will lead to just a single product. with this in mind, predict the product of each of these reactions. (a) (b) no2 (c) br cl? cl2? cl? alcl3 fecl3 alcl3
With this in mind, the product of each of these reactions.(a) NO2, (b) Cl2, and (c) Br. AlCl3 and FeCl3 are the catalysts in each case.
Nitration of Benzene, when benzene is nitrated with a mixture of nitric and sulfuric acid, the nitro group (-NO2) replaces one of the hydrogens on the benzene ring to produce nitrobenzene. When benzene is nitrated with a mixture of nitric and sulfuric acid, the nitro group (-NO2) replaces one of the hydrogens on the benzene ring to produce nitrobenzene. Chlorination of Benzene, when benzene is chlorinated, a hydrogen atom is replaced by a chlorine atom, and the product produced is chlorobenzene
When benzene is chlorinated, a hydrogen atom is replaced by a chlorine atom, and the product produced is chlorobenzene. Bromination of Benzene, bromination of benzene results in the formation of bromobenzene when iron (III) bromide is used as a catalyst. Bromination of benzene results in the formation of bromobenzene when iron (III) bromide is used as a catalyst.
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What is the hydroxide ion concentration and the pH for a hydrochloric acid solution that has a hydronium ion concentration of 1.50 × 10-3 M?
A) 6.67 × 10-12 M, 11.17
B) 6.67 × 10-11 M, 3.82
C) 6.67 × 10-12 M, 2.82
D) 6.67 × 10-11 M, 10.18
The hydroxide ion concentration and the pH for a hydrochloric acid solution that has a hydronium ion concentration of [tex]1.50* 10^{3}[/tex] M are (D) [tex]6.67* 10^{11}[/tex] M, or[tex]10^{18}[/tex]
What is the hydroxide ion?
The hydrogen ion, or hydrogen oxide, is a negatively charged molecule or ion consisting of oxygen and hydrogen atoms with the formula OH. The hydroxide ion is a basic anion with many useful applications in chemistry, such as acid-base reactions and synthesis reactions.
Arrhenius acid is a chemical that donates a proton, whereas Arrhenius base is a chemical that accepts a proton.
The compound created when an acid donates a proton to a base is referred to as a conjugate base, while the compound created when a base accepts a proton from an acid is referred to as a conjugate acid.
When the concentration of H3O+ is given in a solution, it is used to calculate the pH.
The pH scale is used to measure how acidic or basic a solution is. As a result, the pH of the hydrochloric acid solution is calculated as follows:
pH = -log [H3O+] pH = -log (1.50–10) pH = 2.82
We can use the ion-product constant for water to calculate the hydroxide ion concentration.
The equilibrium constant, also known as the ion product constant for water, is equal to the product of the hydronium ion and hydroxide ion concentrations.
[OH][H3O+] = 1.00 x 1014 mol/L2. [OH] = Kw/[H3O+] [OH] = 1.00 x [tex]10^{14}[/tex]mol/L2 / 1.50 10-3 [OH−] = 6.67* 10^{11}
Therefore, the hydroxide ion concentration and the pH for a hydrochloric acid solution that has a hydronium ion concentration of 1.50 [tex]10 ^{3}[/tex]M are6.67* 10^{11} M, or 10^{18}
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how long would it take to plate 8.96 g of nickel onto copper using a 2.0 v power supply with a 315 ma current flow and 0.50 m nickel (ii) acetate solution? assume a current efficiency of 100% and that your plating solution contains the ni(ii) ion.
Using Faraday's law, we can calculate the amount of time it would take to plate 8.96 g of nickel onto copper using a 2.0 V power supply with a 315 mA current flow and a 0.50 M nickel (ii) acetate solution. Faraday's law states that the amount of charge required to plate one mole of metal is equal to the current multiplied by the time.
In this case, we can calculate the time by dividing the total charge (8.96 g/63.55 g/mol = 0.14 mol) by the current (315 mA = 0.315 A). This gives us a total time of 0.45 minutes, or 27 seconds. This assumes a 100% current efficiency, which is often not the case due to the presence of side reactions in the plating solution. Therefore, it is likely that the actual time to plate 8.96 g of nickel will be longer than 27 seconds.
In conclusion, it would take 27 seconds to plate 8.96 g of nickel onto copper using a 2.0 V power supply with a 315 mA current flow and a 0.50 M nickel (ii) acetate solution, assuming 100% current efficiency. However, due to side reactions in the plating solution, the actual time may be longer.
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A 4.1 g sample of gold (specific heat capacity = 0.130 J/g °C) is heated using 52.2 J of energy. If the original temperature of the gold is 25.0°C, what is its final temperature?
The final temperature of the gold is approximately 34.7°C.
To solve this problem, we need to use the formula:
Q = m x c x ΔT
Where Q is the amount of energy absorbed by the gold, m is the mass of the gold, c is the specific heat capacity of gold, and ΔT is the change in temperature.
We are given that the mass of the gold is 4.1 g, the specific heat capacity of gold is 0.130 J/g °C, and the amount of energy absorbed by the gold is 52.2 J. We are also given the initial temperature of the gold, which is 25.0°C.
We can rearrange the formula to solve for ΔT:
ΔT = Q / (m x c)
Plugging in the values, we get:
ΔT = 52.2 J / (4.1 g x 0.130 J/g °C)
ΔT = 99.23 °C
This tells us that the gold has undergone a temperature change of 99.23°C. To find the final temperature, we add this change to the initial temperature:
Final temperature = 25.0°C + 99.23°C
Final temperature = 124.23°C
Therefore, the final temperature of the gold is 124.23°C.
To calculate the final temperature of the gold sample, you can use the formula:
Q = mcΔT
where Q is the energy (52.2 J), m is the mass (4.1 g), c is the specific heat capacity (0.130 J/g°C), and ΔT is the change in temperature (final temperature - initial temperature).
Rearrange the formula to find the final temperature:
ΔT = Q / (mc)
ΔT = 52.2 J / (4.1 g * 0.130 J/g°C)
Now, calculate ΔT:
ΔT ≈ 9.7°C
The initial temperature of the gold is 25.0°C, so the final temperature will be:
Final temperature = Initial temperature + ΔT
Final temperature ≈ 25.0°C + 9.7°C
Final temperature ≈ 34.7°C
Therefore, The gold's ultimate temperature is around 34.7°C.
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How does the type of bond determine the naming convention for compounds?
The type of bond between atoms in a compound determines the naming convention for the compound.
Compounds can be broadly classified into two categories:ionic compounds and covalent compounds. In ionic compounds, the bond between the atoms is ionic, which means that one or more electrons are transferred from one atom to another. The resulting ions are held together by electrostatic forces, which creates a strong bond. In naming ionic compounds, the cation (positively charged ion) is listed first, followed by the anion (negatively charged ion).
In covalent compounds, the bond between the atoms is covalent, which means that electrons are shared between the atoms. Covalent compounds can be further classified into polar and nonpolar compounds. In naming covalent compounds, the elements are listed in order of increasing electronegativity, and prefixes are used to indicate the number of atoms of each element in the compound.
Therefore, the type of bond in a compound determines the naming convention used to name the compound.
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Calculate the molarity of the sodium bicarbonate solution. Also recall that 1 tsp = 4. 9mL qnd 1 cup = 236. 6 mL
The molarity of the sodium bicarbonate solution is 11.2 M.
We need to know the amount of sodium bicarbonate (NaHCO₃) dissolved in a given volume of solution to calculate its molarity. Let's assume that we have dissolved 1 tsp (teaspoon) of NaHCO₃ in a solution.
According to the problem, 1 tsp = 4.9 mL. We need to convert the volume into liters:
4.9 mL = 4.9/1000 L = 0.0049 L
Now we need to know the amount of NaHCO₃ in moles that is dissolved in this volume of solution. The molar mass of NaHCO₃ will be 84.01 g/mol. Let's assume that the NaHCO₃ is pure and completely dissolved in the solution.
We can use the following formula to calculate the molarity of the solution:
Molarity (M) = moles of solute/liters of solution
moles of NaHCO₃ = mass of NaHCO₃ / molar mass of NaHCO₃
Assuming 1 tsp of NaHCO₃ is equivalent to 4.6 g (which is the approximate weight of 1 tsp of NaHCO₃), we can calculate the moles of NaHCO₃:
moles of NaHCO₃ = 4.6 g / 84.01 g/mol = 0.0548 moles
Now we substitute the values into formula for molarity:
Molarity (M) = 0.0548 moles / 0.0049 L
= 11.2 M
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a student creates a solution with a molarity of 1.55 m. if the solute has a molar mass of 110 g/mol and the solution contains 188 g of solute, what is the volume of the solution?
the volume of the solution is: 1.1035 L or 1103.5 mL.
The given is as follows:
a student creates a solution with a molarity of 1.55 m.
If the solute has a molar mass of 110 g/mol and
the solution contains 188 g of solute,
what is the volume of the solution?
Let's calculate the volume of the solution. The volume of the solution can be calculated using the formula:
moles of solute = mass of solute ÷ molar mass
molarity = moles of solute ÷ volume of solution
We need to calculate the volume of the solution.
Thus, rearranging the above formula, we get:
Volume of solution = Moles of solute ÷ Molarity
Calculate the number of moles of solute
Number of moles of solute = Mass of solute ÷ Molar mass= 188 g ÷ 110 g/mol= 1.7091 mol
Now, we can calculate the volume of the solution.
Volume of solution = Number of moles of solute ÷ Molarity= 1.7091 mol ÷ 1.55 mol/L≈ 1.1035 L or 1103.5 mL
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What is the three carbon alcohol that forms the backbone of a triglyceride called?
The three carbon alcohol that forms the backbone of a triglyceride is called glycerol.
Glycerol is a type of alcohol that is found in many fats and oils. It is a clear, odorless, viscous liquid that is sweet-tasting and non-toxic. Glycerol is a versatile molecule that is used in a variety of industrial and pharmaceutical applications.
The structure of a triglyceride consists of three fatty acids that are linked to a glycerol molecule. The fatty acids are long-chain hydrocarbons that have a carboxylic acid group (-COOH) at one end. The carboxylic acid group can react with the hydroxyl group (-OH) on the glycerol molecule to form an ester linkage (-COO-). The resulting molecule is a triglyceride. Triglycerides are a type of lipid that are important for energy storage in the body. They are found in adipose tissue and are broken down into fatty acids and glycerol when the body needs to use stored energy.
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The table shows the concentration of a reactant in the reaction mixture over a period of time.
What is the average rate of the reaction over the first 450 seconds?
A. 1.7 × 10−3
B. 1.9 × 10−3
C. 2.0 × 10−3
D. 2.2 × 10−3
Answer:
D
Explanation:
it is 2.2 ×10-3 because the avarage is between 2 and 3
calculate the gradient of the line
The gradient of the line passing through the points (2,3) and (5,7) is 4/3.
The gradient of a line is calculated by dividing the difference in the -coordinates by way of the distinction in the -coordinates. This might also be referred to as the trade in divided by means of the exchange in , or the vertical divided by way of the horizontal.
How do you calculate gradient formula?The gradient equation is another way we refer to the gradient of a straight line the usage of x and y coordinates. So once more the gradient equation is viewed as m = upward jab / run where m is the gradient or slope.
What is the components for gradient of linear graph?Finding the gradient of a straight-line graph
The gradient (also known as slope) of a line passing through two points (x1, y1) and (x2, y2) is given by:
gradient = (y2 - y1) / (x2 - x1)
Using the given points, we can calculate the gradient of the line passing through (2,3) and (5,7) as follows:
gradient = (7 - 3) / (5 - 2)
= 4 / 3
Therefore, the gradient of the line passing through the points (2,3) and (5,7) is 4/3.
The gradient of the line = (change in y-coordinate)/(change in x-coordinate)
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Complete question:
Calculate the gradient of the line passing through the points (2,3) and (5,7).
Translate the given conformer from the wedge-and-dash drawing into its Newman projection.
Select the correct Newman projection using one of the tabs below, and drag the three groups (Cl, Br, and CH3) to their correct locations.
When translating the given conformer from the wedge-and-dash drawing into its Newman projection, there are steps to follow. These steps are explained below:
Step 1: Identify the axial and equatorial atoms in the conformer. Step 2: Determine which of the axial atoms will be in the front and which will be at the back. Step 3: Draw a circle and divide it into four sections. Step 4: Place the front axial atom in the left section of the circle.Step 5: Place the remaining axial atom in the right section of the circle. Step 6: Place the equatorial atom in the bottom section of the circle.Step 7: Rotate the back axial atom by 60 degrees so that it can point upwards.Step 8: Repeat the rotation for the equatorial atom. The final result is the Newman projection.The correct Newman projection can be seen by clicking on the "Tabs" tab. The Cl atom is in the back, while the Br atom is in the front, with the CH3 atom on the right side of the Newman projection.
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experiment 3: suppose you had a buffer containing 0.5 moles of sodium dihydrogen phosphate and 0.5 moles of sodium hydrogen phosphate. how many moles of hydrochloric acid would this phosphate buffer be able to accept before the ph of the solution began to change drastically?
The phosphate buffer can accept 0.5 moles of hydrochloric acid before the pH of the solution begins to change drastically.
In this experiment, you have a buffer containing 0.5 moles of sodium dihydrogen phosphate (NaH2PO4) and 0.5 moles of sodium hydrogen phosphate (Na2HPO4). To determine how many moles of hydrochloric acid (HCl) this phosphate buffer can accept before the pH begins to change drastically, follow these steps:
1. Identify the conjugate acid-base pairs in the buffer system: NaH2PO4 (acid) and Na2HPO4 (base).
2. Calculate the initial moles of both the acid and base in the buffer.
Initial moles of NaH2PO4 = 0.5 moles
Initial moles of Na2HPO4 = 0.5 moles
3. Recognize that when HCl is added, it reacts with the base (Na2HPO4) to form the conjugate acid (NaH2PO4) and NaCl as a byproduct.
HCl + Na2HPO4 -> NaH2PO4 + NaCl
4. Calculate the moles of HCl required to react with all the available base in the buffer.
Since we have 0.5 moles of Na2HPO4, and the reaction occurs in a 1:1 ratio, it would require 0.5 moles of HCl to react with all the available base.
5. Determine the point at which the pH of the buffer begins to change drastically.
This occurs when all the available base (Na2HPO4) has been consumed by the added HCl, and the buffer capacity has been exceeded. At this point, the buffer can no longer maintain a constant pH.
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give an explain why the henderson hasselbalch equation can be used at the beginning and at the equivalence point of the titration
The Henderson-Hasselbalch equation can be used at the beginning and at the equivalence point of the titration because it relates the pH of a solution to the ratio of the concentration of the conjugate base and the weak acid of a buffer system.
At the beginning of a titration, the buffer system is intact, and the concentration ratio of the weak acid and its conjugate base remains constant. Therefore, the pH of the solution can be calculated using the Henderson-Hasselbalch equation.At the equivalence point of a titration, the moles of the acid and base in the solution are equal, and the buffer system is no longer present.
However, the Henderson-Hasselbalch equation can still be used to determine the pH of the solution because it relates the pH to the ratio of the concentration of the acid and the base. At the equivalence point, the concentration of the acid is equal to the concentration of its conjugate base, and the pH of the solution can be calculated using the Henderson-Hasselbalch equation.
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