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2 years ago

1. A volume of 120.0 liters of a gas is prepared at a pressure of 400.0 mm Hg and a temperature

Chemistry

1 answer:

Olegator [25]2 years ago

8 0

Answer:

$V_2=2487.9L$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the final volume by using the combined ideal gas:

$\frac{P_2V_2}{T_2}=\frac{P_1V_1}{T_1}$

In such a way, by solving for the final volume, V2, we obtain:

$V_2=\frac{P_1V_1T_2}{P_2T_1} \\\\$

Now, by plugging in the pressures, temperatures in Kelvins and initial volume, we will obtain:

$V_2=\frac{(400.0mmHg)(120.0L)(283.15K)}{(20.0mmHg)(273.15K)}\\\\V_2=2487.9L$

Regards!

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The equilibrium: 2 NO2(g) \Longleftrightarrow&iff; N2O4(g) has Kc = 4.7 at 100ºC. What is true about the rates of the forwar

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Explanation:

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Because in the initial state there's only NO₂, there's no possibility for the reverse reaction (from N₂O₄ to NO₂). Thus the forward rate will be larger than the reverse rate.

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3 years ago

a 25 ml volume of a sodium hydroxide solution requires 19.6 mL of a 0.189 M HCl acid for neutralization. A 10 mL volume of phosp

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Answer: 0.172 M

Explanation:

a) To calculate theconcentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=0.189M\\V_1=19.6mL\\n_2=1\\M_2=?\\V_2=25mL$

Putting values in above equation, we get:

$1\times 0.189\times 19.6=1\times M_2\times 25\\\\M_2=0.148$

b) To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_3PO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=?\\V_1=10mL\\n_2=1\\M_2=0.148\\V_2=34.9mL$

Putting values in above equation, we get:

$3\times M_1\times 10=1\times 0.148\times 34.9\\\\M_1=0.172M$

The concentration of the phosphoric acid solution is 1.172 M

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3 years ago

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3 years ago

At a certain temperature the vapor pressure of pure benzene is measured to be . Suppose a solution is prepared by mixing of benz

Marianna [84]

Answer:

P(C₆H₆) = 0.2961 atm

Explanation:

I found an exercise pretty similar to this, so i'm gonna use the data of this exercise to show you how to do it, and then, replace your data in the procedure so you can have an accurate result:

<em>"At a certain temperature the vapor pressure of pure benzene (C6H6) is measured to be 0.63 atm. Suppose a solution is prepared by mixing 79.2 g of benzene and 115. g of heptane (C7H16) Calculate the partial pressure of benzene vapor above this solution. Round your answer to 2 significant digits. Note for advanced students: you may assume the solution is ideal".</em>

<em />

Now, according to the data, we want partial pressure of benzene, so we need to use Raoul's law which is:

P = Xₐ * P° (1)

Where:

P: Partial pressure

Xₐ: molar fraction

P°: Vapour pressure

We only have the vapour pressure of benzene in the mixture. We need to determine the molar fraction first. To do this, we need the moles of each compound in the mixture.

To get the moles: n = m / MM

To get the molar mass of benzene (C₆H₆) and heptane (C₇H₁₆), we need the atomic weights of Carbon and hydrogen, which are 12 g/mol and 1 g/mol:

MM(C₆H₆) = (12*6) + (6*1) = 78 g/mol

MM(C₇H₁₆) = (7*12) + (16*1) = 100 g/mol

Let's determine the moles of each compound:

moles (C₆H₆) = 79.2 / 78 = 1.02 moles

moles (C₇H₁₆) = 115 / 100 = 1.15 moles

moles in solution = 1.02 + 1.15 = 2.17 moles

To get the molar fractions, we use the following expression:

Xₐ = moles(C₆H₆) / moles in solution

Xₐ = 1.02 / 2.17 = 0.47

Finally, the partial pressure is:

P(C₆H₆) = 0.47 * 0.63

Hope this helps

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3 years ago