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Natasha2012 [34]
1 year ago
13

What volume would a 0. 250 mole sample of h2 gas occupy if it had a which a a pressure of 1. 70 atm and a temperature of 35 C?

Chemistry
1 answer:
frosja888 [35]1 year ago
4 0

The volume of  0. 250 mole sample of H_{2} gas occupy if it had a pressure of 1. 70 atm and a temperature of 35 °C is  3.71 L.

Calculation,

According to ideal gas equation which is known as ideal gas law,

PV =n RT

  • P is the pressure of the hydrogen gas  = 1.7 atm
  • Vis the volume of the hydrogen gas = ?
  • n is the number of the hydrogen gas = 0.25 mole
  • R is the universal gas constant = 0.082 atm L/mole K
  • T is the temperature of the sample = 35°C = 35 + 273 = 308 K

By putting all the values of the given data like pressure temperature universal gas constant and number of moles in equation (i) we get ,

1.7 atm×V = 0.25 mole ×0.082 × 208 K

V = 0.25 mole ×0.082atm L /mole K × 308 K /1.7 atm

V = 3.71 L

So, volume of the sample of the hydrogen gas occupy is  3.71 L.

learn more about ideal gas equation

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photosynthesis, is an endothermic reaction in which energy is absorbed from the surrounding.

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Calculate the number of grams of sulfuric acid in 1 gallon of battery acid if the solution has a density of 1.25 g/mL and is 37.
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4 0
2 years ago
Which of the following elements matches the criteria below?
Tems11 [23]

Answer:

The answer to your question is below

Explanation:

1. Found in period 2. All the elements in the list are found in period 2.

a. F   This option is correct

b. Be  Beryllium is located in period two.

c. O  also oxygen is found in period 2.

d. C Carbon is found in period 2.

2.- Can gain lose 4 electrons to become its nearest stable noble gas. Only Carbon.

a. F    This option is wrong, F becomes stable when it gains 1 electron.

b. Be  Beryllium becomes stable when it loses 2 electrons.

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<u>d. C </u><u>Become stable when it gains or loses 4 electrons.</u>

4 0
2 years ago
How many g of silver chloride will be produced by reacting 10g of silver nitrate with sodium chloride?
I am Lyosha [343]

Answer:

16.8 g of AgCl are produced

Explanation:

The reactants are: NaCl and AgNO₃

The products are:  AgCl, NaNO₃

Balanced equation:  NaCl(aq) +  AgNO₃(aq)  → NaNO₃(aq) + AgCl(s) ↓

We convert the mass of AgNO₃ to moles → 10 g / 85g/mol = 0.117 moles

Ratio is 1:1, therefore 0.117 moles of nitrate will produce 0.117 moles of AgCl.

According to stoichiormetry.

We convert the moles to mass → 0.117 mol . 143.3g /1mol = 16.8 g

7 0
3 years ago
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Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing
Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of AgNO_3 = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of AgNO_3 = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and AgNO_3.

\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles

\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag

From the balanced reaction we conclude that

As, 2 mole of AgNO_3 react with 1 mole of Zn

So, 0.1479 moles of AgNO_3 react with \frac{0.1479}{2}=0.07395 moles of Zn

From this we conclude that, Zn is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Ag

From the reaction, we conclude that

As, 2 mole of AgNO_3 react to give 2 mole of Ag

So, 0.1479 moles of AgNO_3 react to give 0.1479 moles of Ag

Now we have to calculate the mass of Ag

\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag

\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g

Theoretical yield of Ag = 15.95 g

Experimental yield of Ag = 13.32 g

Now we have to calculate the percent yield.

\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100

\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%

Therefore, the percent yield is, 83.51 %

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