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kiruha [24]
3 years ago
15

PLEASE ANSWER THIS HURRY

Chemistry
1 answer:
Scrat [10]3 years ago
3 0
Its a covalent bond for this q
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Which does not affect chemical equilibrium
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What is the molality of a solution in which 0.42 moles aluminum chloride has been dissolved in 4200 water ?
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Answer:

0.1 M

<h3>Explanation:</h3>
  • Molarity refers to the concentration of a solution in moles per liter.
  • It is calculated by dividing the number of moles of solute by the volume of solvent;
  • Molarity = Moles of the solute ÷ Volume of the solvent

<u>In this case, we are given;</u>

  • Number of moles of the solute, NH₄Cl as 0.42 moles
  • Volume of the solvent, water as 4200 mL or 4.2 L

Therefore;

Molarity = 0.42 moles ÷ 4.2 L

            = 0.1 mol/L or 0.1 M

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7 0
3 years ago
The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.
harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})

= 70molCO_2

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})

= 87.5molO_2

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

84.0molO_2(\frac{4molO_2}{5molO_2})

= 67.2molCO_2

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