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Andrej [43]
3 years ago
10

Write the balance equation of ammonium sulfide + aluminum chloride ​

Chemistry
1 answer:
Liono4ka [1.6K]3 years ago
5 0

Answer:

3(NH4)2S + 2AlCl3 → 6NH4Cl + Al2S3

Explanation:

Firstly, ammonium sulfide has a chemical formula of (NH4)2S while aluminum chloride has a chemical formula of AlCl3. These two chemical substances react as follows:

(NH4)2S + AlCl3 → NH4Cl + Al2S3

This chemical equation is however, unbalanced. To balance the chemical equation, one has to ensure that the atoms of each element on both the reactant and product side equates with one another.

If we carefully observe each atom on both the reactants and product side, we will have the following balanced equation:

3(NH4)2S + 2AlCl3 → 6NH4Cl + Al2S3

That is;

- 6 moles of N and Cl on both sides

- 24 moles of H on both sides

- 3 moles of S on both sides

- 2 moles of Al on both sides

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What is biodiversity? I don't really understand...
Vesnalui [34]

Answer:

it is also know as biological diversity

Explanation:

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4 0
3 years ago
How many moles are there in 35.2g of Gold?
ki77a [65]

Explanation:

196.96655

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5 0
3 years ago
2Al + 6HCl --&gt; 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L
dem82 [27]

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

<em>Moles H2:</em>

PV = nRT; PV/RT = n

<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

<em />

<em>Moles Al:</em>

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

<em />

<em>Mass Al -Molar mass: 26.98g/mol-:</em>

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries

7 0
3 years ago
Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t
Liono4ka [1.6K]

Answer: The value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

\Delta T_{b} = 3.80^{o}C

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m

Now, the values of K_b is calculated as follows.

\Delta T_{b} = i\times K_{b} \times m

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

K_{b} = molal boiling point elevation constant

Substitute the values into above formula as follows.

\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m

Thus, we can conclude that the value of K_{b} for chloroform is 3.62^{o}C/m when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0
3 years ago
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