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chubhunter [2.5K]
3 years ago
6

What is the molarity of a solution made with 400 grams of copper II chloride in 3.5 liters of water?

Chemistry
1 answer:
Leni [432]3 years ago
4 0

Answer:

0.85M

Explanation:

Given parameters:

Mass of CuCl₂  = 400g

Volume of water  = 3.5L

Unknown:

Molarity of the solution = ?

Solution:

Molarity is one of the ways of expressing the concentrations of a solution. It is defined as the number of moles per unit volume of a solvent.

          Molarity  = \frac{number of moles}{volume}

To find the number of moles of CuCl₂;

    Number of moles  = \frac{mass}{molar mass}

  Molar mass of CuCl₂ = 63.6 + 2(35.5)  = 134.6g/mol

   Number of moles  = \frac{400}{134.6}   = 2.97moles

      Molarity  = \frac{2.97}{3.5}    = 0.85M

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0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidised. ... Thus, 90g of lower oxide contains as much metal as 100g of higher oxide, i.e., 80g (given). Hence, 80g of metal combines with 10g of oxygen in the lower oxide and 20g of oxygen in the higher oxide.

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If 4.65 LL of CO2CO2 gas at 22 ∘C∘C at 793 mmHg mmHg is used, what is the final volume, in liters, of the gas at 35 ∘C∘C and a p
otez555 [7]

Answer:

About 7.9 L.

Explanation:

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\displaystyle PV = nRT

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Because the right-hand side stays constant, we have that:
\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = nR

Hence substitute initial values and known final values:
\displaystyle \begin{aligned} \frac{(793\text{ mm Hg})(4.65 \text{ L})}{(22 \text{ $^\circ$C})} & = \frac{(743 \text{ mm Hg})V_2}{(35\text{ $^\circ$C})} \\ \\ V_2 & = 7.9\text{ L}\end{aligned}

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8 0
2 years ago
The following solutions are prepared by dissolving the requisite amount of solute in water to obtain the desired concentrations.
lions [1.4K]

Answer:

1M MgCl₂ > 1M KCl > 1M C₁₂H₂₂O₁₁

Explanation:

The osmotic pressure (π) is the pressure needed to impede the osmose, it means that it's the necessary pressure to prevent the solvent to go through a membrane.

It can be calculated by:

π = M*R*T*i

Where M is the molarity of the solution (mol/L), R is the ideal gas constant, T is the temperature, and i the van't Hoff factor.

This factor is a way to correct the number of particles that are dissolved in a solute, and it can be calculated by:

i = 1 + α*(q - 1)

Where α is the degree of dissociation of a substance, and q is the number of moles of each ion released in a solution. Thus, covalent compounds that didn't ionize, such as sugars, have only one particle, and q = 1, and so i =1.

Because all the substances have the same molarity (1 M) and are at the same temperature, let's analyze the value of i, which is directly proportional to π.

C₁₂H₂₂O₁₁ is a sugar that didn't ionize, so π = 1;

Both KCl and MgCl₂ are soluble salts and will dissociate completely (α = 1), but MgCl₂ will have 3 particles (Mg²⁺ + 2Cl⁻), and KCl only one particle (K⁺ and Cl⁻), so qMgCl₂ > 1KCl, and so πMgCl₂ > πKCl, which will be higher than 1.

1M MgCl₂ > 1M KCl > 1M C₁₂H₂₂O₁₁

5 0
3 years ago
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