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2 years ago

Which describes a way to speed up the collisions between hydrogen and oxygen molecules to produce more water?

2 answers:

4 0

<span>Equation:2H2(g) + O2(g) → 2H2O(g)
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Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.

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mariarad [96]2 years ago

4 0

<h2>Equation:2H2(g) + O2(g) → 2H2O(g)</h2>

<h3>Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.</h3>

You might be interested in

(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g

Anarel [89]

Answer:

Part A

The volume of the gaseous product is $V = 787L$

Part B

The volume of the the engine’s gaseous exhaust is $V_e = 2178 \ L$

Explanation:

Part A

From the question we are told that

The temperature is $T = 350^oC = 350 +273 =623K$

The pressure is $P = 735 \ torr = \frac{735}{760} = 0.967\ atm$

The of $C_8 H_{18} = 100.0g$

The chemical equation for this combustion is

$2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}$

The number of moles of $C_8 H_{18}$ that reacted is mathematically represented as

$n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }$

The molar mass of $C_8 H_{18}$ is constant value which is

$M = 114.23 \ g/mole$

So $n = \frac{100 }{114.23} }$

$n = 0.8754 \ moles$

The gaseous product in the reaction is $CO_2_{(g)}$ and water vapour

Now from the reaction

2 moles of $C_8 H_{18}$ will react with 25 moles of $O_2$ to give (16 + 18) moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

1 mole of $C_8 H_{18}$ will react with 12.5 moles of $O_2$ to give 17 moles of $CO_2_{(g)}$ and $H_2 O_{(g)}$

This implies that

0.8754 moles of $C_8 H_{18}$ will react with (12.5 * 0.8754 ) moles of $O_2$ to give (17 * 0.8754) of $CO_2_{(g)}$ and $H_2 O_{(g)}$

So the no of moles of gaseous product is

$N_g = 17 * 0.8754$

$N_g = 14.88 \ moles$

From the ideal gas law

$PV = N_gRT$

making V the subject

$V = \frac{N_gRT}{P}$

Where R is the gas constant with a value $R = 0.08206 \ L\cdot atm /K \cdot mole$

Substituting values

$V = \frac{14.88* 0.08206 *623}{0.967}$

$V = 787L$

Part B

From the reaction the number of moles of oxygen that reacted is

$N_o = 0.8754 * 12.5$

$N_o = 10.94 \ moles$

The volume is

$V_o = \frac{10.94 * 0.08206 *623}{0.967}$

$V_o = 579 \ L$

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

$V_e = V_o * \frac{0.79}{0.21}$

Substituting values

$V_e = 579 * \frac{0.79}{0.21}$

$V_e = 2178 \ L$

3 0

3 years ago

Which of the following is true about the kinetic molecular theory?

velikii [3]

Answer: (B) Pressure is due to the collisions of the gas particles with the walls of the container.

Option B helps to explain the factor behind gas collision under high pressure.

Explanation: Kinetic molecular theory explains the behaviour and movement of gas particles when they are in motion. It states that gas particles are always in continuous motion and are perfectly elastic in nature.

Kinetic molecular theory can be explained using both Boyle's law and Charles's law.

•Few Assumptions of Kinetic Molecular Theory.

1. Gas particles are always in motion and they collide with the walls of their container.

2. The space occupied by a gas particles is negligible in comparison to the volume of the gas

8 0

3 years ago

Using your knowledge of chemistry and the information in Reference Table H, which statement concerning propanone and water at 50

stiv31 [10]

The answer is (2) higher vapor pressure and weaker intermolecular forces. Propanone has a lower boiling point, so it is more volatile than water. Propanone's vapor pressure is, therefore, higher than that of water at 50 degrees Celsius. Propanone is more volatile due to the fact that the intermolecular forces that hold its molecuels together are not as strong as those that hold together molecules of water. Since the IMFs are weaker, it takes less thermal energy to break individual molecules free of each other.

3 0

3 years ago

Water has a K, value of<br> 1x 10-13<br> 1x 10-10<br> 1x 10-15<br> 1x 10-14

Margaret [11]

K value of water is 1x 10^-14

3 0

2 years ago

30 its ASAP!!!! What is the volume, in liters, occupied by 0.485 moles of Oxygen gas at 23.0 oC and 0.980 atm?

USPshnik [31]

Use the universal gas formula

PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = <span>0.08205 L atm / (mol·K)

Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)

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5 0

3 years ago