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kompoz [17]
3 years ago
11

Please help, will give brainliest

Chemistry
1 answer:
sleet_krkn [62]3 years ago
5 0

Answer:

Mass = 61.6 g

Explanation:

Given data:

Mass of ammonia produced = 75.0 g

Mass of nitrogen needed = ?

Solution:

Chemical equation:

N₂ + 3H₂    →      2NH₃

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 75 g/ 17 g/mol

Number of moles =  4.4 mol

No we will compare the moles of ammonia with nitrogen.

              NH₃           :          N₂

                2             :           1

               4.4           :          1/2×4.4 =2.2 mol

Mass of N₂:

Mass = number of moles × molar mass

Mass = 2.2 mol × 28 g/mol

Mass = 61.6 g

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 <span>H2SO4 gives 2 moles oh H+ per mole of acid 

[H2SO4] = 2M so [H+] = 4M 

pH = -log(4) = -.6 

Therefore, the pH </span><span>of a 2.0 M H2SO4 solution is -0.6
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4 years ago
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You have 2 moles of NaOH how many molecules of NaOH?
kaheart [24]

Answer:

12.044 X 10^23 molecules of NaOH

Explanation:

because NaOH is an ionic bond, we should be asking how many <em>formula units </em> are in 2 moles of NaOH, there are 0 molecules since NaOH is measured in formula units.

but for the sake of the problem I'll assume NaOH is measured in molecules

for every mol of something there are 6.022 X 10^23 of something of that something.

so there are 6.022 X 10^23 molecules for every mol of NaOH

that means we have 2 X 6.022 X 10^23 molecules in 2 moles of NaOH = 12.044 X 10^23 molecules of NaOH

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What is the procedure used to separate nitrogen from the mixture of nitrogen and oxygen​
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Answer:

Cryogenic distillation

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How can the amount of electrical energy produced be less than the initial radiant energy from the sun?
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2 years ago
The sulfur dioxide (so2) stack-gas concentration from fossil-fuel combustion is 12 ppmv. determine the stack-gas so2 concentrati
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The concentration of SO_{2} in the stack gas = 12 ppmv

That means 12 L of  SO_{2} is present per 10^{6} L gas

The given temperature is 273 K (0 C) and pressure is 1 atm. At these conditions, 1 mol of gas would occupy,

PV = nRT

(1 atm) (V) = (1 mol)(0.08206\frac{L.atm}{mol.K}) (273 K)

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10^{6} L *\frac{1 m^{3}}{1000 L} = 10^{3}   m^{3}

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