First, we need to write the balanced chemical equation for the reaction:
2 NO2(g) + H2O(l) → HNO3(aq) + NO(g)
From the equation, we can see that 2 moles of NO2 react with 1 mole of H2O to produce 1 mole of HNO3 and 1 mole of NO. Therefore, we need to determine which reactant is limiting and calculate the amount of HNO3 that can be produced based on that.
To do this, we can use the mole ratio of NO2 to H2O:
5.0 mol NO2 × (1 mol H2O / 2 mol NO2) = 2.5 mol H2O
Since we have 11.0 mol of H2O, it is not limiting and we will use up all of the NO2.
Therefore, we can calculate the amount of HNO3 that can be produced from 5.0 mol of NO2:
5.0 mol NO2 × (1 mol HNO3 / 2 mol NO2) = 2.5 mol HNO3
Therefore, the largest amount of HNO3 that could be produced is 2.5 mol, rounded to the nearest 0.1 mol.
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If 80 grams of KBr were dissolved in 100 grams of water at 35 degrees Celsius, which of these terms would best describe the solution
The term that would best describe the solution formed if 80g KBr dissolved in 100g water is unsaturated solution.
What is a saturated solution?A saturated solution is a solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom.
On the other hand, an unsaturated solution is that solution that is capable of dissolving more of a solute at the same temperature.
According to this question, 80 grams of KBr were dissolved in 100 grams of water at 35 degrees Celsius. This means that the solution is unsaturated because it can still dissolve more KBr.
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What’s the oxidation number of copper in CuO?
the oxidation number of copper in copper oxide is 2...
Answer: +2
Explanation: Copper has a +2 oxidation number in CuO.
This is due to the fact that oxygen has an oxidation number of 2, and the entire chemical has a neutral charge. Consequently, the following equation can be used to determine copper's oxidation number:
(+2) + (-2) = 0
In order to counteract the -2 oxidation number of oxygen in CuO, copper must have an oxidation number of +2.
Aqueous sulfuric acid (H₂SO₂) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na₂SO) and liquid water (H₂O). What is the
theoretical yield of sodium sulfate formed from the reaction of 4.9 g of sulfuric acid and 5.0 g of sodium hydroxide?
Round your answer to 2 significant figures.
The theoretical yield of sodium sulfate, Na₂SO₄, formed from the reaction of 4.9 g of sulfuric acid, H₂SO₄ and 5.0 g of sodium hydroxide, NaOH is 7.1 g
How do i determine the theoretical yield?First, we shall determine the limiting reactant. This is shown below:
H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O
Molar mass of H₂SO₄ = 98 g/molMass of H₂SO₄ from the balanced equation = 1 × 98 = 98 g Molar mass of NaOH = 40 g/molMass of NaOH from the balanced equation = 2 × 40 = 80 gFrom the balanced equation above,
98 g of H₂SO₄ reacted with 80 g of NaOH
Therefore,
4.9 g of H₂SO₄ will react with = (4.9 × 80) / 98 = 4 g of NaOH
From the above calculation, we can see that only 4 g of NaOH out of 5 g is needed to react with 4.9 g H₂SO₄.
Thus, the limiting reactant is H₂SO₄
Finally, we shall determine theoretical yield of sodium sulfate, Na₂SO₄ formed. Details below:
H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O
Molar mass of H₂SO₄ = 98 g/molMass of H₂SO₄ from the balanced equation = 1 × 98 = 98 gMolar mass of Na₂SO₄ = 142 g/molMass of Na₂SO₄ from the balanced equation = 1 × 142 = 142 gFrom the balanced equation above,
98 g of H₂SO₄ reacted to produce 142 g of Na₂SO₄
Therefore,
4.9 g of H₂SO₄ will react to produce = (4.9 × 142) / 98 = 7.1 g of Na₂SO₄
Thus, the theoretical yield of sodium sulfate, Na₂SO₄ formed is 7.1 g
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Describe the technique for washing a precipitate. Place the steps in the correct order.
A. add deionized water
B. mix solutions
C. decant
D. centrifuge
What is a solvent front?
Answer:
A solvent front is the point on a chromatography paper or plate where the solvent has reached the end of the stationary phase and has migrated as far as it can go. It is the farthest point reached by the solvent in the chromatography process.
In a reaction between vinegar and antacid tablets, the antacid is the limiting reagent. cm of gas. At constant pressure and temperature, three tablets produce 600 cm³ What volume will four tablets produce? 300 cm³ 600 cm³ 800 cm³ 3 1,200 cm³ 3
If in a reaction between vinegar and antacid tablets, the antacid is the limiting reagent. cm of gas. At constant pressure and temperature, three tablets produce 600 cm³ . The volume that four tablets will produce is: C. 800 cm³.
What volume will four tablets produce?Since the antacid is the limiting reagent, the amount of gas produced will be directly proportional to the number of tablets used.
We know that three tablets produced 600 cm³ of gas. Therefore, we can set up a proportion:
3 tablets produce 600 cm³ of gas
4 tablets produce x cm³ of gas
To solve for x, we can use cross-multiplication:
3 tablets × x cm³ of gas = 4 tablets × 600 cm³ of gas
3x = 2400
x = 800 cm³
Therefore the answer is C. 800 cm³.
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How many hydrogen molecules (h2) are needed to convert the triacylglycerol shown to saturated fat
We would need about 16 hydrogen atoms so that we can convert the compound to a saturated fat.
What is a saturated fat?In animal products like meat and dairy, saturated fat is a form of dietary fat that is normally solid at room temperature. It is known as being "saturated" because each molecule of fat has the most hydrogen atoms possible, giving it a stable structure.
We can see this by counting the number of double bonds in the fat and there are eight of them so sixteen hydrogen atoms are needed for saturation.
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The critical point for water lies at 275 °C and 3.2 atm, calculate the DH°vap of water.
The ΔH°vap of water at the critical point is approximately 0.04614 kJ/mol.
To calculate the ΔH°vap (enthalpy of vaporization) of water at the critical point, we can use the Clausius-Clapeyron equation;
ln(P₂/P₁) = ΔH°vap/R [1/T₁ - 1/T₂]
where P₁ and T₁ are the pressure and temperature at which the enthalpy of vaporization is known (usually at standard conditions of 1 atm and 100 °C), P₂ and T₂ are the pressure and temperature at the critical point, ΔH°vap is the enthalpy of vaporization, and R is the gas constant (8.314 J/mol∙K).
Using the given values, we can plug them into the equation and solve for ΔH°vap;
ln(3.2 atm / 1 atm) = ΔH°vap / R [1/373 K - 1/275 K]
Simplifying;
ln(3.2) = ΔH°vap / R [0.0026819]
ΔH°vap / R = ln(3.2) / 0.0026819
ΔH°vap / R = 5.552
Multiplying both sides by R:
ΔH°vap = 5.552 x R
ΔH°vap = 5.552 x 8.314 J/mol∙K
ΔH°vap = 46.14 J/mol
Converting to kJ/mol;
ΔH°vap = 0.04614 kJ/mol
Therefore, the ΔH°vap of water is 0.04614 kJ/mol.
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The question is in the picture
Charles's law of gases states that the density of an ideal gas is inversely proportional to its temperature at constant pressure.
The equation is as follows;
Va/Ta = Vb/Tb
Where;
Va and Ta = initial volume and temperature respectivelyVb and Tb = final volume and temperature respectively0.67/362 = 1.12/Tb
0.00185Tb = 1.12
Tb = 605.41K
This temperature in °C is 605.41 - 273 = 332°C
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Predict which of the following reactions has a positive change in entropy.
I. 2N2(g) + O2(g) → 2N2O(g)
II. CaCO3(s) → CaO(s) + CO2(g)
III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Answer:
Explanation:
The change in entropy of a system can be determined by comparing the entropy of the reactants to the entropy of the products. The reaction that leads to an increase in the number of moles of gas or particles will generally have a positive change in entropy.
I. 2N2(g) + O2(g) → 2N2O(g)
The reactants have 3 moles of gas, while the product also has 3 moles of gas. Therefore, there is no change in the number of moles of gas, and the change in entropy is likely to be small.
II. CaCO3(s) → CaO(s) + CO2(g)
The reactant is a solid, while the products are a solid and a gas. The formation of a gas from a solid leads to an increase in the number of moles of particles, and therefore an increase in entropy.
III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
The reactants consist of a solid and a liquid, while the products consist of an aqueous solution and a gas. The formation of a gas leads to an increase in the number of moles of particles, and therefore an increase in entropy.
Therefore, reactions II and III have a positive change in entropyentropy
Question 6 (5 points)
Label each situation as metals, nonmetals, or metalloids. Label each with numbers.
Usually conducts electricity
& heat well
at room temperature these
are gases or liquids
Will lose valance electrons
to form compounds.
can be used as
semiconductors
Will gain valance electrons
to form compounds.
1. a metal
2. a nonmetal
3. a metalloid
Answer:
Explanation:
Here are the labels for each situation:
1.Usually conducts electricity & heat well - Metal (1)
2.At room temperature these are gases or liquids - Nonmetal (2)
3.Will lose valance electrons to form compounds - Metal (1)
4.Can be used as semiconductors - Metalloid (3)
5.Will gain valance electrons to form compounds - Nonmetal (2)
Write the cations and anions present in CrO2
The chemical molecule CrO2 is also known as chromium(IV) oxide or chromic acid. It has the molecular formula CrO2 and is an inorganic substance.
In the solid state, CrO2 exists as a solid with a layered structure, and it is considered a cationic compound. The cation present in CrO2 is chromium(IV) ion, denoted as Cr4+.
On the other hand, the anion present in CrO2 is oxide ion, denoted as O2-. The oxidation state of oxygen in this compound is -2.
So, the cations present in CrO2 are Cr 4+ ions, and the anions present are O2 -2 ions.
In CrO2, the cation present is Chromium (Cr) with a charge of +4, and the anion present is Oxygen (O) with a charge of -2.
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25 points and I’ll mark as brainliest!!! Tasks are in the picture.
Answer:
5. 0.566 g
6. A. 100 times more
Explanation:
5. The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration. For a solution with pH=2, the concentration of hydrogen ions is 10^-2 mol/L. Since HBr is a strong acid, it dissociates completely in water to produce H+ and Br- ions. Therefore, the concentration of HBr in the solution is also 10^-2 mol/L.
The molar mass of HBr is 80.91194 g/mol
So, in a 700 mL solution (0.7 L), there are
0.7 L * 10^-2 mol/L = 0.007 mol of HBr.
This corresponds to 0.007 mol * 80.91194 g/mol = 0.566 g of HBr dissolved in the solution.
6. The pH of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration. This means that for each decrease in pH by 1 unit, the hydrogen ion concentration increases by a factor of 10. Since the difference in pH between the two solutions is 3 units (6-3=3), the hydrogen ion concentration in the solution with pH=3 is 10^3 = 100 times more than in the solution with pH=6.
1. To operate a batch reactor for converting A into R. This is a liquid phase reaction with the stoichiometry A → R. CA,(mol/l) 0.1 0.2 0.3 0.4 0.2 0.6 0.7 0.8 1.0 1.3 2.0 -rA,(mol/l min) 0.1 0.3 0.5 0.6 0.5 0.25 0.10 0.06 0.05 0.045 0.042 For the above data determine the order of reaction and rate constant.
The reaction is second order with a rate constant of 0.043 mol/l min.
How to explain the reactionFor CA = 0.1 mol/l, -rA = 0.1 mol/l min
For CA = 0.2 mol/l, -rA = 0.3 mol/l min
For CA = 0.3 mol/l, -rA = 0.5 mol/l min
For CA = 0.4 mol/l, -rA = 0.6 mol/l min
The slope of this line is equal to the order of the reaction (n), and the y-intercept is ln(k).
Slope = (0.6931 - (-2.3026)) / (0.3010 - (-0.9163)) = 1.929
ln(k) = -2.3026 + 1.929 * (-0.3010)
ln(k) = -3.1504
k = e^(-3.1504) = 0.043 mol/l min
The reaction is second order with a rate constant of 0.043 mol/l min.
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The 500 cm³ of a pas enclosed in a container under a pressure of 580 mm of Hg. If the volume is reduced to 300 cm³ what will be the pressure then?
Answer:
The answer is 966.67 mm of Hg.
Explanation:
To solve this problem, we can use Boyle's Law which states that the pressure of a gas is inversely proportional to its volume when the temperature and the amount of gas are kept constant. The formula for Boyle's Law is:
P1V1 = P2V2
where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.
Using the given values:
P1 = 580 mmHg
V1 = 500 cm³
V2 = 300 cm³
We can solve for P2:
P1V1 = P2V2
580 mmHg x 500 cm³ = P2 x 300 cm³
290,000 mmHg·cm³ = P2 x 300 cm³
P2 = 290,000 mmHg·cm³ / 300 cm³
P2 = 966.67 mmHg (rounded to the nearest hundredth)
Therefore, the pressure when the volume is reduced to 300 cm³ is approximately 966.67 mmHg.
Which solution would you choose to supress the dissolution of MgCO3?
A. 0.200 M NaCl
B. 0.200 HCl
C. 0.200 M NaNO3
D. 0.200 M Na2CO3
The best solution to suppress the dissolution of MgCO3 is option D 0.200 M Na2CO3
To suppress the dissolution of MgCO3We need to add an ion or compound that will react with MgCO3 and form a precipitate, thus removing Mg2+ and CO32- ions from the solution.
Therefore, Option D, 0.200 M Na2CO3, contains CO32- ions that can react with Mg2+ ions to form MgCO3 precipitate. This would effectively suppress the dissolution of MgCO3 by removing Mg2+ and CO32- ions from the solution.
Therefore, option D is the best solution to suppress the dissolution of MgCO3.
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How does the presence of coal in Antarctica support Wegener's continental drift hypothesis?
Answer:
Explanation:
Coal deposits have been found in Antarctica, particularly in the Transantarctic Mountains where they are interbedded in sedimentary rocks of the flat-lying Beacon Supergroup. The presence of coal in Antarctica supports Wegener's continental drift hypothesis because it suggests that Antarctica was once part of a larger landmass that had a warm climate suitable for the formation of coal. Coal is formed from ancient plant matter that has been compressed and heated over millions of years. The presence of coal in Antarctica suggests that the continent was once located closer to the equator and had a climate that supported lush vegetation.
Question: Why is the liquid oxygen machine producing less liquid oxygen than normal?
Claim1: there is frozen water in tank 2, which is blocking some of the oxygen from coming into tank 3.
Claim2: some of the liquid oxygen evaporated in tank 3.
Claim3: some of the oxygen didn’t condense in tank 2.
Explain why group 8 elements of the periodic table are referred to as group 0
Answer: They have eight outer electrons.
Explanation: They don't need any more electrons to be added to them and can't give out any electrons to other groups. They have a complete outer shell.
HELP PLEASE
A 6.50-g sample of copper metal at 25.0 °C is heated by the addition of 145 J of energy. The final temperature of the copper is ________ °C. The specific heat capacity of copper is 0.38 J/g-K.
58.7
33.7
83.7
25.0
33.5
A 6.50-g sample of copper metal at 25.0 °C is heated by the addition of 145 J of energy. The final temperature of the copper is 83.7 °C. The specific heat capacity of copper is 0.38 J/g-K.
The correct answer choice is "83.7"
To solve this problem, we can use the equation:
q = mcΔT
where q is the amount of energy absorbed by the copper, m is the mass of the copper, c is the specific heat capacity of copper, and ΔT is the change in temperature of the copper.
Rearranging this equation to solve for ΔT, we get:
ΔT = q / (mc)
Substituting the given values, we get:
ΔT = 145 J / (6.50 g x 0.38 J/g-K)
ΔT = 58.7 K
Therefore, the final temperature of the copper is:
25.0 °C + 58.7 °C = 83.7 °C
So the correct option is 83.7.
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A chemistry teacher has 6 liters of a
sodium nitrate solution. She has 24
students in her class and she wants
to divide the solution evenly among
them. How many milliliters of sodium
nitrate solution will each student
receive?
Answer:
There are 1000 milliliters (ml) in one liter. Therefore, the teacher has a total of 6 x 1000 = 6000 ml of sodium nitrate solution.
Explanation:
To divide the solution evenly among the 24 students, we need to divide the total volume of the solution by the number of students:
6000 ml ÷ 24 students = 250 ml per student
Therefore, each student will receive 250 milliliters of sodium nitrate solution.
Answer:
Answer- 0.25ml
Explanation:
So there are 24 students and 6 liters of Solution.So to evenly distribute
Just divide 6 by 24(6÷24/)... So the answer will be 0.25
How many g Al must react with iodine to form AlI₃ via the following reaction scheme to release -836.0 kJ of heat? 2 Al(s) + 3 I₂(s) → 2 AlI₃(s)
∆H = -302.9 kJ
The mass (in grams) of aluminum, Al that must react with iodine to form AlI₃, given that -836.0 KJ of heat is relaesd is 149.0 g
How do i determine the mass aluminum required?The mass of aluminum required to react with iodine to produce AlI₃ can be obtain as shown below:
2Al(s) + 3I₂(s) → 2AlI₃(s) ∆H = -302.9 KJ
Molar mass of aluminum, Al = 27 g/molMass of aluminum, Al from the balanced equation = 2 × 27 = 54 gFrom the balanced equation above,
When -302.9 KJ of heat energy is released, 54 g of aluminum, Al reacted.
Therefore,
When -836.0 KJ of heat energy will be release = (-836.0KJ × 54 g) / -302.9 KJ = 149.0 g of aluminum, Al will react.
Thus, from the above calculation, we can conclude that the mass of aluminum, Al required is 149.0 g
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Here are some data from a similar experiment, to determine the empirical formula of on oxide of tin.
Calculate the empirical formula according to these data.
Mass of crucible, cover, and tin sample 21.76 g
Mass of empty crucible with cover 19.66 g
Mass of crucible and cover and sample,
after prolonged heating gives constant weight 22.29 g
The information given can be used to construct the empirical formula for a tin oxide. We must first determine the mass of tin in the sample. This may be achieved by deducting the mass of the crucible, cover, and sample (21.76 g) from the mass of the empty crucible and cover (19.66 g).
This gives us a mass of 2.10 g of tin in the sample. The mass of oxygen in the sample must then be determined. To achieve this, we must deduct the mass of the crucible, cover, and sample (21.76 g) from the mass of the same components (22.29 g) prior to protracted heating. This provides us with an oxygen mass of 0.53 g.
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What mass (grams) of sodium sulfate would be formed by the complete reaction of 137.3 grams of sodium hydroxide?
NaOH + H2SO4 --> Na2SO4 + HOH
The mass of sodium sulfate [tex](Na_2SO_4)[/tex] formed by the complete reaction of 137.3 grams of sodium hydroxide [tex](NaOH)[/tex] is 486.74 grams.
The balanced chemical equation for the reaction between sodium hydroxide [tex](NaOH)[/tex] and sulfuric acid [tex](H_2SO_4)[/tex] is:
[tex]NaOH[/tex] + [tex](H_2SO_4)[/tex] → [tex]Na_2SO_4[/tex] + [tex]2H_2O[/tex]
From the balanced equation, we can see that 1 mole of [tex]NaOH[/tex] reacts with 1 mole of [tex](H_2SO_4)[/tex] to produce 1 mole of [tex]Na_2SO_4[/tex] and 2 moles of water [tex](H_2O).[/tex]
The molar mass of [tex]NaOH[/tex] is 40.00 g/mol, which means that 137.3 grams of [tex]NaOH[/tex] is equal to 137.3 g / 40.00 g/mol = 3.4325 moles of [tex]NaOH[/tex].
Since 1 mole of [tex]NaOH[/tex] reacts with 1 mole of [tex](H_2SO_4)[/tex] to produce 1 mole of [tex](Na_2SO_4)[/tex] , we can say that 3.4325 moles of [tex]NaOH[/tex] will react with 3.4325 moles of [tex]H_2SO_4[/tex] to produce 3.4325 moles of [tex]Na_2SO_4[/tex].
The molar mass of [tex](Na_2SO_4)[/tex] is 142.04 g/mol, which means that 1 mole of [tex](Na_2SO_4)[/tex] has a mass of 142.04 g.
Therefore, 3.4325 moles of [tex](Na_2SO_4)[/tex] has a mass of 3.4325 moles x 142.04 g/mol = 486.74 grams.
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Chemistry what is the reaction Rate TABLE
The rate constant, k = 5.27 E-2 s⁻¹, determines the rate law for the reaction P → E + Z.
How to determine rate constant?The rate of the reaction P → E + Z can be expressed as:
Rate = - d[P]/dt = d[E]/dt = d[Z]/dt
where d[P], d[E], and d[Z] = changes in the concentrations of P, E, and Z, respectively, over a small time interval dt.
Use the experimental data to determine the rate constant and the order of the reaction.
Calculate the initial rate of the reaction in each trial by dividing the change in concentration of P by the time interval:
rate1 = (d[P]/dt)1 = (0.30 M - 0 M)/(20 s) = 0.015 M/s
rate2 = (d[P]/dt)2 = (0.60 M - 0.30 M)/(20 s) = 0.015 M/s
rate3 = (d[P]/dt)3 = (0.90 M - 0.60 M)/(20 s) = 0.015 M/s
The initial rates are the same in all three trials, which suggests that the reaction is first-order with respect to P.
Now using any of the three trials to determine the value of the rate constant k, trial 1:
Rate1 = k[P]1
k = Rate1/[P]1 = (1.58 E-2 M/s)/(0.30 M) = 5.27 E-2 s⁻¹
Therefore, the rate law for the reaction P → E + Z is:
Rate = k[P]
where k = 5.27 E-2 s⁻¹ is the rate constant.
Use the rate law to calculate the expected rates of the reaction at different concentrations of P. For example:
Rate2 = k[P]2 = (5.27 E-2 s⁻¹)(0.60 M) = 3.16 E-2 M/s
Rate3 = k[P]3 = (5.27 E-2 s⁻¹)(0.90 M) = 4.74 E-2 M/s
These expected rates are close to the experimental rates, which suggests that the rate law is a good approximation for the reaction under these conditions.
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Chemistry. . . Reaction: AB₂C (g) → B₂ (g) + AC (g), find the value of K
At equilibrium [AB₂C]=0.0168 M, [B₂]= 0.007 M, and [AC] = 0.0118 M
The value of K at equilibrium, for the reaction is 0.0049
How do i determine the value of K at equilibrium?First, we shall list out the given parameters from the question. This is shown below:
AB₂C (g) ⇌ B₂(g) + AC(g) Concentration of AB₂C, [AB₂C] = 0.0168 MConcentration of B₂, [B₂]= 0.007 MConcentration of AC, [AC] = 0.0118 MEquilibrium constant (K) =?Equilibrium constant is defined as:
Equilibrium constant = [Product]ᵐ / [Reactant]ⁿ
Where
m is the coefficient of productsn is the coefficient of reactantsWith the above formula, we can obtain the equilibrium constant, K as follow:
Equilibrium constant, K = [B₂][AC] / [AB₂C]
K = (0.007 × 0.0118) / 0.0168
K = 0.0049
Thus, the equilibrium constant, K for the reaction is 0.0049
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CHEMISTRY chemistry Table balance A+B→C Table2
Answer:
zn + Hcl cual es su rreaccion
What is the molar volume of CO2 at 39 C and 652 torr?
The molar volume of a gas can be calculated using the ideal gas law:
PV = nRT
where P is the pressure of the gas in atmospheres (atm), V is the volume of the gas in liters (L), n is the number of moles of gas, R is the ideal gas constant (0.08206 L·atm/mol·K), and T is the temperature of the gas in Kelvin (K).
To solve for the molar volume of CO2 at 39°C (312 K) and 652 torr (0.859 atm), we can rearrange the ideal gas law as follows:
V = (nRT) / P
First, we need to calculate the number of moles of CO2. We can use the following equation, which relates the pressure, volume, number of moles, and temperature of a gas:
PV = nRT
Solving for n, we get:
n = (PV) / (RT)
Substituting the given values, we get:
n = (0.859 atm * V) / (0.08206 L·atm/mol·K * 312 K)
Now we can substitute this expression for n into the equation for the molar volume:
V = (nRT) / P
V = [(0.859 atm * V) / (0.08206 L·atm/mol·K * 312 K)] * (0.08206 L·atm/mol·K * 312 K) / (0.859 atm)
Simplifying, we get:
V = 24.45 L/mol
Therefore, the molar volume of CO2 at 39°C and 652 torr is 24.45 L/mol.
Oxygen and oxygen-containing compounds are involved in many different reactions. Which of the following equations represent a balanced reaction involving 14 atoms of oxygen? Question 5 options: NH4Cl + KOH --> NH3 + H2O + KCl 2Na + 2H2O --> 2NaOH + H2 2C2H6 + 7O2 --> 4CO2 + 6H2O 4Fe + 3O2 --> 2Fe2O3
The equation that represents a balanced reaction involving 14 atoms of oxygen is:
2C2H6 + 7O2 --> 4CO2 + 6H2O
a. The relationship between variables in this equation is that 2 moles of C2H6 (ethane) react with 7 moles of O2 (oxygen) to produce 4 moles of CO2 (carbon dioxide) and 6 moles of H2O (water). This equation follows the law of conservation of mass, where the total number of atoms on both sides of the equation is the same, indicating a balanced reaction.
b. The graph is linear, as the coefficients of the reactants and products in the equation are whole numbers and form a consistent ratio. The coefficients of 2, 7, 4, and 6 represent the stoichiometry of the reaction, indicating a fixed relationship between the reactants and products.
c. An example of a situation where this balanced equation could be applicable is the combustion of ethane (C2H6) in the presence of excess oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), which is a common reaction in the combustion of hydrocarbon fuels. The equation represents the balanced stoichiometry of this reaction, where 2 moles of ethane react with 7 moles of oxygen to produce 4 moles of carbon dioxide and 6 moles of water, involving a total of 14 atoms of oxygen in the reaction.
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1. Which of the following stars has a temperature of approximately 9000 K and luminosity about to
20 times greater than the Surfs luminos
a Sirius
b. Procyon
c. Figel
d. Polaris
2. Which of the following types of stars is considered part of the main sequera
a Supergants
b. Red giants
c. Red dwarts
d. White dwarfs
3. Which of the following stars is cooler than the
Surf
a. Procyon B
b. Pigel
C. Barnard's Star
d. Sirius
4. The Sun is classified with which of the following types of stars?
a. Supergiants
b. Red giants
c. Main sequence
d. White dwars
5. Which of the forces listed below is most responsible for the formation of start?
a. Gravity
b. Magnetism
c. Bectromagnetism
d. Light
6. Which star has a higher luminosity and a lower temperature than the Sun?
a. Pigel
b. Barnard's Star
c. Alpha Centauri
d. Aldebaran
7. Compared to the temperature and luminosity of the star Polars, the star Srus is
a. hotter and more luminous
b. hotter and less luminous
c. cooler and more luminous cooler and less luminous
1. The star that has a temperature of approximately 9000 K and luminosity about 20 times greater than the Sun’s luminosity is Vega.
2. The type of star that is considered part of the main sequence is red dwarfs.
3. The star that is cooler than the Sun is Barnard’s Star.
4. The Sun is classified as a main sequence star.
5. The force most responsible for the formation of stars is gravity.
6. The star that has a higher luminosity and a lower temperature than the Sun is Aldebaran.
7. Compared to the temperature and luminosity of the star Polaris, the star Sirius is hotter and more luminous.