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stepan [7]
2 years ago
10

How many grams of CaF2 are present in 1.25 L of a 0.15 M solution of CaF2? How do I find the grams I am confused on that part?

Chemistry
1 answer:
dybincka [34]2 years ago
7 0

Answer:

Mass = 14.64 g

Explanation:

Given data:

Volume of solution = 1.25 L

Molarity of Solution = 0.15 M

Mass of CaF₂ = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

We will calculate the number of moles of CaF₂ and then determine the mass by using number of moles.

0.15 M =  number of moles of solute / 1.25 L

number of moles of solute = 0.15 M ×  1.25 L

number of moles of solute = 0.1875 mol/L × L

number of moles of solute = 0.1875 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.1875 mol ×78.07 g/mol

Mass = 14.64 g

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A bird building a nest for its young is a reproductive strategy. Based on this example, what do you think the term “reproductive
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Acetaldehyde decomposes at 750 K: CH3CHO → CO + CH4. The reaction is first order in acetaldehyde and the half-life of the reacti
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1. Calculate the molarity of a sugar solution if 4 liters of the solution contains 8 moles of sugar
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Explanation:

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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in
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Explanation:

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        m = 10.0 kg = 10,000 g    (as 1 kg = 1000 g)

      Initial temp. of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K  

      Initial temp. of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

            mC \Delta T = mC \times \Delta T

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                   2T_{f} = 100^{o}C

                          T_{f} = 50^{o}C

Convert temperature into kelvin as (50 + 273) K = 323 K.              

Also, we know that the relation between enthalpy and temperature change is as follows.

             \Delta H = mC \Delta T

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                         = 1243550 J

or,                     = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

     \Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

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            = 10000 g \times 0.385 J/K g \times -0.143

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   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

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Thus, we can conclude that for the given reaction \Delta H is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

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