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evablogger [386]
3 years ago
8

A mining crew extracted two different types of minerals from underground.

Chemistry
2 answers:
Alexus [3.1K]3 years ago
7 0

they could be different types of minerals, so one reacted to the energy differently than the other one. and pls answer my latest questions its due by 11:59 pm TONIGHT or i get kicked out of school.. pls help..

Oksana_A [137]3 years ago
4 0

Answer:

Mineral A's molecules had a high level of attraction before. As well as the molecules in mineral B.  But, mineral A's molecules, probably had higher kinetic energy. Which means a higher temperature and faster-moving molecules. Which allowed it to spread more and since it spread more, it became a liquid.  Mineral B might have had not that much kinetic energy so the temperature was colder and its molecules moved slower. That's why probably mineral A changed to a liquid while mineral B remained a Solid.

Explanation:

Mineral A's molecules had a high level of attraction before. As well as the molecules in mineral B.  But, mineral A's molecules, probably had higher kinetic energy. Which means a higher temperature and faster-moving molecules. Which allowed it to spread more and since it spread more, it became a liquid.  Mineral B might have had not that much kinetic energy so the temperature was colder and its molecules moved slower. That's why probably mineral A changed to a liquid while mineral B remained a Solid.

(Answer + Explanation are both the same thing btw)

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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
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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

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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

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