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umka21 [38]
3 years ago
13

If you start with 6 moles of N2 and 6 moles of H2 (meaning you won't have enough of 1 of the ingredients), how many moles of NH3

would you be able to make?
Chemistry
1 answer:
Ivahew [28]3 years ago
8 0

Answer:

4molNH_3

Explanation:

Hello there!

In this case, according to the given information it will be firstly necessary to set up the chemical equation taking place:

N_2+3H_3\rightarrow 2NH_3

We infer we need to calculate the moles of NH3 by using both of the moles of N2 and H2 at the beginning, in order to identify the limiting reactant:

n_{NH_3}=6molN_2*\frac{2molNH_3}{1molN_2}=12molNH_3\\\\ n_{NH_3}=6molH_2*\frac{2molNH_3}{3molH_2}=4molNH_3\\

Thus, since hydrogen yields the fewest moles of ammonia, we conclude that we are just able to yield 4 moles of NH3.

Regards!

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Calculate the concentration of clo2− at equilibrium if the initial concentration of hclo2 is 2. 25×10^−2 m.
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The concentration of ClO₂⁻ at equilibrium if the initial concentration of  HClO₂ is 0.0654.

<h3>What is concentration?</h3>

The concentration of any substance is the quantity of that substance in per square of the space or container.

The reaction is

HClO₂ + H₂O <=> H₃O⁺ + ClO₂⁻

The pH is 0.454 M

Ka = [H₃O⁺][ClO₂⁻ ] / [HClO₂]

2. 25 × 10⁻² m = [x][x] / 0.454-x]

2 + 0.011 - 0.004994 = 0

solve the quadratic equation

x = 0.0654 = [H3O+] = [ClO2-]

pH = -log (H3O+)

pH = -log(0.0654)

pH = 1.2

equilibrium concentrations of

[HClO2] = 0.454 -x = 0.454 -0.0654 = 0.3886 M

[ClO2- ] = x = 0.0654

Thus, the equilibrium concentrations  is 0.0654.

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6 0
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Which is an example of a physical change?
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Answer:

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Explanation:

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Given that w for water is 2. 4×10−14 m^2 at 37 °C. Calculate the ph of a neutral aqueous solution at 37 °c, which is the normal
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The pH of a neutral aqueous solution at 37°C is 6.8.

<h3>What is Kw? </h3>

Kw is defined as the dissociation, which is also known as self-ionization, constant of water. this is an equilibrium constant, and its expression is:

Kw = [OH⁻] . [H₃O⁺]

Neutral pH determines that the concentrations of OH⁻ and H₃O⁺ are equal.

<h3>Calculation</h3>

Let us suppose concentration of OH and H₃O⁺ is x, to calculate it:

Kw =[OH⁻] . [H₃O⁺] = x²

x² = 2.4 × 10⁻¹⁴ M²

x = 1.5919 × 10⁻⁷ M

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pH = -log[H₃O⁺] = -log( 1.5919×10⁻⁷ M)

pH = 6.8

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