If the bark is stripped off a tree, the tree will

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t

myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = $n_3$

$n=n_1+n_2+n_3$

$2.22=0.5 mol +0.6 mol+ n_3$

$n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol$

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

3 0

3 years ago

You have a graduated cylinder with 10 mL of water in it.

Law Incorporation [45]

Answer:

<h3>The answer is 10 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 300 g

volume = final volume of water - initial volume of water

volume = 40 - 10 = 30 mL

We have

$density = \frac{300}{30} = \frac{30}{3} \\$

We have the final answer as

Hope this helps you

5 0

3 years ago

What is the net Ionic equation of HClO4(aq)+ NaOH(aq) = H2O(l)+ NaClO4(aq)

uranmaximum [27]

H^+(aq) + OH^-(aq) ---> H2O(l)

<span>Na^+ and ClO4^- are the spectator ions.</span>

7 0

3 years ago

Will mark as brainiest <br> Please help

Levart [38]

Answer:

metal sulfate
metal sulfate
copper sulfate
copper nitrate
copper chloride
copper phosphate
hydrochloric acid, water
Potassium, sulfuric acid, water

(Correct me if I am wrong)

4 0

3 years ago

Read 2 more answers

Balance the equation <br><br> __C7H6O2 + __O2 —> __CO2 + __H2O

ser-zykov [4K]

2 C₇H₆O₂ + 15 O₂ → 14 CO₂ + 6 H₂O

<u>Explanation:</u>

C₇H₆O₂ + O₂ → CO₂ + H₂O

First we have to balance the O- atoms, we have to put 6 in front of water so there are 12 H atoms on RHS, to balance it we need to put 2 in front of C₇H₆O₂, and so we have 14 C - atoms on LHS, 28 + 6 = 34 O - atoms on RHS, so we have to put 15 in front of Oxygen in LHS, so that each and every atom in the equation gets balanced now. The balanced equation is, 2 C₇H₆O₂ + 15 O₂ → 14 CO₂ + 6 H₂O

4 0

3 years ago