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bazaltina [42]
1 year ago
13

How many moles of solute would be dissolved in .500 kg of solvent to make a 2.50 molal NaOH solution?

Chemistry
1 answer:
Tom [10]1 year ago
3 0

Molality is a measure of concentration that relates the moles of solute to the kilograms of solvent, it is described by the following equation:

Molality=\frac{molSolute}{kgSolvent}

We are given the molality(2.50m) and kilograms of solvent(0.500kg), so we solve for moles of solute from the equation:

molSolute=Molality\times kgSolvent\begin{gathered} molSolute=2.50m\times0.500kg \\ molSolute=1.25molSolute \end{gathered}

To make a 2.50molal NaOH solution would be needed 1.25moles of solute

Answer: 1.25 moles

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Answer:

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Explanation:

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6 0
3 years ago
Enter your answer in the provided box. Calculate the number of moles of CrCl, that could be produced from 49.4 g Cr202 according
Mrrafil [7]

Answer:

0.4694 moles of CrCl₃

Explanation:

The balanced equation is:

Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)

The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.

The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:

MCr = 52 g/mol

MCl = 35.5 g/mol

MO = 16 g/mol

So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.

The number of moles is the mass divided by the molar mass, so:

n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.

For the stoichiometry:

1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃

0.2347 mol of Cr₂O₃----------- x

By a simple direct three rule:

x = 0.4694 moles of CrCl₃

6 0
3 years ago
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vodomira [7]
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5 0
3 years ago
4NH3 + 3O2 --> 2N2 + 6H2O
larisa86 [58]

The balanced reaction is:

4NH3 + 3O2 --> 2N2 + 6H2O

 <span>We are given the amount of reactants to be used for the reaction. This will be the starting point of our calculation.</span>

83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2

2.81 moles of NH3

From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.

<span>2.62 mol O2</span><span> (6 mol H2O  / 3 mol O2) (18.02 g H2O / 1 mol H2O) = 94.42 g H2O</span>

8 0
2 years ago
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