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Agata [3.3K]
3 years ago
7

Ammonia (NH3) reacts with sulfuric acid to form ammonium sulfate. How many grams of ammonium sulfate are obtained from 8.73 g of

ammonia, assuming a 73% yield of product?

Chemistry
1 answer:
Gwar [14]3 years ago
4 0
I hope you understand my working:
1) Finding the mol of NH3 to find the mol of (NH4)2SO4 (ammonium sulfate)
2) Mr of (NH4)2SO4
3) Theoretical yield: The actual grams of (NH4)2SO4 produced when reacting 0.514 mol of NH3 to 0.514 mol H2SO4
4) Using formula of (given grams)/(theoretical grams or actual grams) * 100 = 73%
5) Basic algebra

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A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. What is the weight/
lara31 [8.8K]

Answer:

A solution is made by dissolving 4.87 g of potassium nitrate in water to a final volume of 86.4 mL solution. The weight/weight % or percent by mass of the solute is :

<u>2.67%</u>

Explanation:

Note : Look at the density of potassium nitrate in water if given in the question.

<u><em>You are calculating </em></u><u><em>weight /Volume</em></u><u><em> not weight/weight % or percent by mass of the solute</em></u>

Here the <u>weight/weight % or percent by mass</u> of the solute is asked : So first convert the<u> VOLUME OF SOLUTION into MASS</u>

Density of potassium nitrate in water KNO3 = 2.11 g/mL

density=\frac{mass}{volume}

Density = 2.11 g/mL

Volume of solution = 86.4 mL

2.11=\frac{mass}{86.4}

mass = 2.11\times 86.4

mass=182.3grams

Mass of Solute = 4.87 g

Mass of Solution = 183.2 g

w/w% of the solute =

= \frac{mass\ of\ solute}{mass\ of\ solution}\times 100

=\frac{4.87}{183.2}\times 100

w/w%=2.67%

8 0
3 years ago
What is the pH of a 2.0 x 10^-4 M solution of nitric acid (HNO3)
Semmy [17]

Hello!

datos:

Molarity = 2.0*10^{-4}\:M\:(mol/L)


ps: The ionization constant of the nitric acid is strong (100% ionized in water) or completely dissociates in water, so the pH will be:

pH = - log\:[H_3O^+]

pH = - log\:[2*10^{-4}]

pH = 4 - log\:2

pH = 4 - 0.30

\boxed{\boxed{pH = 3.70}}\end{array}}\qquad\checkmark

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR!

3 0
3 years ago
Read 2 more answers
38.25 grams of silicon is combined with 14.33 grams of nitrogen gas. How many grams of silicon nitride can be formed if nitrogen
zhuklara [117]
3Si + 2N2 --> Si3N4 (as given) 

n(Si) = m/MM = 38.25/28.085 = 1.3619 mol
n(N2) = 14.33/2*14.007 = 0.5115 mol

Therefore, N2 is limiting and Si is in excess 
The molar ratio of 2N2:Si3N4 is 2:1 
So, 0.0575 mol of silicon nitride is formed (dividing 0.5115 by 2) 

m of silicon nitride= n*mm = 0.0575*140.283 = 8.06627... g 
= 8.066g (4 significant figures) 

(hopefully it is right, but double check in case i did something wrong) :) 
6 0
3 years ago
A certain reaction with an activation energy of 205 kj/mol was run at 485 k and again at 505 k . what is the ratio of f at the h
lapo4ka [179]
Arrhenius' Law relates activation energy, Ea, rate constant, K, and temperature, T as per this equation:

K (T) = A * e ^ (-Ea / RT), where R is the universal constant of gases and A is a constant which accounts for collision frequency..

Then you can find the ration between K's at two different temperatures as:

K1 = A * e ^ (-Ea / RT1)

K2 = A* e ^(-Ea / RT2)

=> K1 / K2 = e ^ { (-Ea / RT1) - Ea / RT2) }

=> K1 / K2 = e ^ {(-Ea/ R ) *( 1 / T1 - 1 T2) }

=> K1 / K2 = e^ { (-205,000 j/mol / 8.314 j/mol*k )* ( 1 / 505K - 1/ 485K) }

=> K1 / K2 = e ^ (2.0134494) ≈ 7.5

Answer: 7.5




8 0
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AURORKA [14]

Answer: yes

Explanation: no

5 0
2 years ago
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