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8090 [49]
2 years ago
13

Correct first answer= Brianliest!PLEASE WOULD REALLY APPRECIATE IF YOU’D HELP ME OUT:’(

Chemistry
1 answer:
sveticcg [70]2 years ago
3 0

Answer:

The answer to your question is: 2 grams of methane

Explanation:

Data

V = 9.15 l

P = 1.77 atm

T = 57° C = 330 °K

Ar mass = 19 g

CH4 mass = ?

Formula

PV = nRT

Process

                     n = \frac{PV}{RT}

                      n = \frac{(1.77)(9.15)}{(0.082)(330)}

                      n = \frac{16.26}{27.06}

                             n = 0.6

Argon

                     40 g of Ar -------------------- 1 mol

                      19 g         ---------------------   x

                                 x = 0.475 mol of Ar

moles of CH4 = 0.6 - 0.475

                       = 0.125

                     

Methane

                            16 g of CH4 ---------------  1mol

                             x                 ----------------  0.125 mol

                            x = 2 grams

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Which of the following would be expected to have the highest viscosity? CH3CH2OH, HOCH2OH, CH3CH2CH3
zaharov [31]

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Explanation:

The resistance occurred in the flow of a liquid substance is called viscosity.

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For example,  HOCH_{2}OH has strong intermolecular hydrogen bonding than the one's present in CH_{3}CH_{2}OH and CH_{3}CH_{2}CH_{3}. This is because two-OH groups are present over here.

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4 0
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A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr
sertanlavr [38]

Answer: The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

Explanation:

The combined gas equation is,

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

where,

P_1 = initial pressure of gas = 0.947 atm

P_2 = final pressure of gas = 0.987 atm

V_1 = initial volume of gas = 150 ml

V_2 = final volume of gas = 144 ml    

T_1 = initial temperature of gas = 25^0C=(25+273.15)K=298.15K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}

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The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

4 0
2 years ago
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