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natka813 [3]
3 years ago
8

What is the volume of 2.5 moles of nitrogen gas (N2)at standard temperature and pressure (STP)?​

Chemistry
1 answer:
gregori [183]3 years ago
8 0

Answer:

1 mole of gas = 22.4L

2.5 moles of gas takes up = ( 22.4 L/ 1 mole ) x 2.5 mole

= 56 L

Explanation:

Please mark brainliest and have a great day!

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A eudiometer contains a 65.0 ml sample of a gas collected
SCORPION-xisa [38]

Answer:

53.1 mL

Explanation:

Let's assume an ideal gas, and at the Standard Temperature and Pressure are equal to 273 K and 101.325 kPa.

For the ideal gas law:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is temperature, 1 is the initial state and 2 the final state.

At the eudiometer, there is a mixture between the gas and the water vapor, thus, the total pressure is the sum of the partial pressure of the components. The pressure of the gas is:

P1 = 92.5 - 2.8 = 89.7 kPa

T1 = 23°C + 273 = 296 K

89.7*65/296 = 101.325*V2/273

101.325V2 = 5377.45

V2 = 53.1 mL

6 0
3 years ago
Which of the following is not an example of a chemical change?
ale4655 [162]
Salt dissolving in a glass of water, using electricity to break down water into hydrogen and oxygen, rust forming on an iron fence and gas burning on a stove.  
4 0
3 years ago
Read 2 more answers
Mercury has an atomic mass of 200.59 amu. calculate the mass of 3.0 x 10^10 atoms
Licemer1 [7]
To determine mass of the given number of atoms of mercury, we need a factor that would relate the number of atoms to number of moles. In this case, we use the Avogadro's number. It is a <span>number that represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. The number of units could be atoms, molecules, ions or electrons. To convert into mass, we use the given amu of mercury since it is equal to grams per mole. We calculate as follows:

</span>3.0 x 10^10 atoms ( 1 mol / 6.022 x 10^23 atoms ) ( 200.59 g / 1 mol ) = 9.99x10^-12 g Hg
5 0
3 years ago
. Element X has a nucleon (mass) number of 19 and a proton (atomic) number of 9.
Marina86 [1]

Answer:

C

Explanation:

it belong to that group as it needs 1 electron to be chemically stable

7 0
3 years ago
Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:
Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: Al_{2}S_{3} +H_{2}O  _\to  Al(OH)_{3} + H_{2}S

First, balance the chemical equation

Al_{2}S_{3} + 6H_{2}O  \to 2Al(OH)_{3} + 3H_{2}S

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

n_{Al_{2}S_{3}  } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3}  } = \frac{158g}{150.16g/mol}   \\n_{Al_{2}S_{3} }=1.05 mol

From the chemical reaction , the ratio of molar is 3mol H_{2}S/1 mol Al_{2}S_{3}. So, the moles of hydrogen sulfide are:

n_{H_{2} O} =\frac{131g}{18.02g/mol}

       = 7.26mol

From the chemical reaction, the molar ratio is 3 mol H_{2}S/6 mol H_{2}O. So, the moles of hydrogen sulfide are:

Moles of H_{2}S formed = 7.26 mol H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }

Th liming reactant isAl_{2}S_{3} beacuse the mass of Al_{2}S_{3} forms less product than water. Therefore, the maximum number of moles of H_{2}S is 3.15 mol.  We know that molar mass of H_{2}S is 34.10g/mol. So, the maximum mass of H_{2}S formed is,

m_{H_{2}S } = n_{H_{2}S } \times Molar mass of H_{2}S

         = 3.15 mol \times 34.10g/mol

         = 107.4g

Now, multiplying the number of moles of Al_{2}S_{3} by the molar ratio between Al_2S_3 and H_2O which is 6mol H_2O/1mol Al_2S_3 we get the number of moles of H_2O reacted.

Moles of H_2O reacted = 1.05 mol Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}

                                     = 6.31 mol H_2O

The mass of H_2O is,

m_{H_{2} O} = 6.31 mol \times 18.02g/ mol

          = 114g

On subtracting, the mass of H_2O reacted from the given mass of H_2O is,

m_{H_2O} = (131-114)g

         = 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

#SPJ4

 

8 0
2 years ago
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