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enyata [817]
3 years ago
8

Ammonia (NH3) breaks down into nitrogen and water by: 4NH3 + 3O2 > 2N2 + 6H2O. How many miles of water will be produced if 12

moles of ammonia decomposes?
Chemistry
1 answer:
Allushta [10]3 years ago
6 0

Answer:

\boxed{ \text{18 mol}}

Explanation:

(a) Balanced equation

4NH₃ + 3O₂ ⟶ 2N₂ + 6H₂O

(b) Calculation

You want to convert moles of NH₃  to moles of H₂O

The molar ratio is 6 mol H₂O:4 mol NH₃

\text{Moles of H$_{2}$O} =\text{12 mol NH$_{3}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{4 mol NH$_{3}$}} = \textbf{18 mol H$_{2}$O}\\\\\text{The reaction will produce }\boxed{ \textbf{18 mol of H$_{2}$O}}

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0.45

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7 0
3 years ago
Lithium acetate, LiCH3CO2, is a salt formed from the neutralization of the weak acid acetic acid, CH3CO2H, with the strong base
Vesna [10]

Answer : The pH of 0.289 M solution of lithium acetate at 25^oC is 9.1

Explanation :

First we have to calculate the value of K_b.

As we know that,

K_a\times K_b=K_w

where,

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K_w = dissociation constant of water = 1\times 10^{-14}

Now put all the given values in the above expression, we get the dissociation constant of a base.

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Now we have to calculate the concentration of hydroxide ion.

Formula used :

[OH^-]=(K_b\times C)^{\frac{1}{2}}

where,

C is the concentration of solution.

Now put all the given values in this formula, we get:

[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}

[OH^-]=1.3\times 10^{-5}M

Now we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (1.3\times 10^{-5})

pOH=4.9

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1

Therefore, the pH of 0.289 M solution of lithium acetate at 25^oC is 9.1

4 0
4 years ago
If the crucible originally weighs 3.715 g and 2. 000 g of hydrate are added to it , what is the weight of the water that is lost
dimulka [17.4K]
For this, you need to know 1) the mass of the hydrate and 2) the mass of the anhydrous salt. Once you have both of these, you will subtract 1) from 2) to find the mass of the water lost.

From the problem, you know that 1) = 2.000 g.

Now you need to find 2). You know that your crucible+anhydrous salt is 5.022 g. To find just the anhydrous salt, subtract the mass of the crucible (3.715 g).

1) = 5.022 g - 3.715 g = 1.307 g

Now you can complete our original task.

Mass H2O = 2) - 1) = 2.000 g - 1.307 g = 0.693 g.


4 0
4 years ago
A 2.10g of unknown monoprotic acid is titrated with 38.10 mL of .265M NaOH. Calculate the molar mass of the ac? I already calcul
Shkiper50 [21]
V_{NaOH}=38,1mL=0,0381L\\
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HR    +      NaOH ⇒ NaR + H₂O
1mol   :     1mol

0,0101mol \ \ \ \ \Rightarrow \ \ \  \ 2,1g\\
1mol  \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ m\\\\
m=\frac{1mol*2,1g}{0,0101mol}\approx 207,92g\\\\
M_{HR}=207,92\frac{g}{mol}
5 0
3 years ago
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