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kondor19780726 [428]

3 years ago

11

1. If 3.2 L of nitrogen monoxide and 4.3 g of oxygen gas are combined, what volume of nitrogen dioxide will be formed? 2NO + 0,

-> 2NO,

1 answer:

Free_Kalibri [48]3 years ago

3 0

Answer:

$3.2 \div 22.4 = \frac{1}{7} mol \: no$

$4.3 \div 32 = 0.134mol \: o2$

Explanation:

NO is limit reactant so:

1÷7×46=6.57 grams of NO2 will form

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Question 2 (1 point)

Alexxandr [17]

Answer:

a mole have have wrong becouse i understand

4 0

3 years ago

Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at

MariettaO [177]

Answer:

P_2 =0.51 atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}$

$0.008389261745 \times 273 = 4.5P_2$

$P_2 =\dfrac{0.008389261745 \times 273 }{4.5}$

$P_2 =0.51 \ atm$

4 0

3 years ago

Marianna [84]

Answer:

B

Explanation:

Trust me I've had problems like these

6 0

3 years ago

Read 2 more answers

Lesson 14.2 Interpret graph boyles law

docker41 [41]

As what we can see on the graph of the Boyle's Law, we can imply that volume and pressure are inversely proportional. The gas law furthermore explains that at this condition, the temperature must be held constant. The law can be furthermore be explained using the equation:

PV = k

6 0

3 years ago

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e

sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$

<u>Initial:</u> 1.8 4.7

<u>At eqllm:</u> 1.8-x 4.7-x

Evaluating the value of 'x'

$\Rightarrow (1.8-x)=1.16\\\\x=0.64$

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}$

$p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm$

Putting values in above expression, we get:

$K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136$

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$

<u>Initial:</u> 2.36 4.06 0.64

<u>At eqllm:</u> 2.36-x 4.06-x 0.64+x

Putting value in the equilibrium constant expression, we get:

$0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41$

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

3 0

3 years ago