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Blababa [14]
3 years ago
7

A 25.225 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 75.815 g of wa

ter. A 14.842 g aliquot of this solution is then titrated with 0.1068 M HCl . It required 29.99 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste
Chemistry
1 answer:
Rudik [331]3 years ago
6 0

Answer:

1.43 (w/w %)

Explanation:

HCl reacts with NH3 as follows:

HCl + NH3 → NH4+ + Cl-

<em>1 mole of HCl reacts per mole of ammonia.</em>

Mass of NH3 is obtained as follows:

<em>Moles HCl:</em>

0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>

<em>Mass NH3 in the aliquot:</em>

3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.

Mass of sample + water = 22.225g + 75.815g = 98.04g

Dilution factor: 98.04g / 14.842g = 6.6056

That means mass of NH3 in the sample is:

0.0545g * 6.6056 = 0.36g NH3

Weight percent is:

0.36g NH3 / 25.225g * 100

<h3>1.43 (w/w %)</h3>
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3 years ago
Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies at 800˚C (1073 K) is 3.6 ×
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Explanation :

Formula used :

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N_v=[\frac{N_A\times \rho}{M}]\times e^{(\frac{-E}{K\times T})}

where,

N_v = equilibrium number of vacancies = 3.6\times 10^{23}m^{-3}=3.6\times 10^{20}L^{-1}

E = energy = ?

M = atomic weight = 107.9 g/mole

N_A = Avogadro's number = 6.022\times 10^{23}mol^{-1}

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

3.6\times 10^{20}L^{-1}=[\frac{(6.022\times 10^{23}mol^{-1})\times 9500g/L}{107.9g/mol}]\times e^{[\frac{-E}{(1.38\times 10^{-23}J/K)\times 1073K}]}

E=1.762\times 10^{-19}J

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4 0
3 years ago
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