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tiny-mole [99]
3 years ago
11

A chemist adds 485 mL of a 0.0025 mol/L calcium sulfate solution to a reaction flask. Calculate the mass in grams of calcium sul

fate the chemist has added to the flask. Round your answer to significant digits.
Chemistry
1 answer:
viva [34]3 years ago
6 0

Answer : The mass in grams of calcium sulfate is 0.16 grams.

Explanation :

Molarity : It is defined as the number of moles of solute present in one litre of solution.

Formula used :

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution}}

Solute is, CaSO_4

Given:

Molarity of CaSO_4 = 0.0025 mol/L

Molar mass of CaSO_4 = 136 g/mole

Volume of solution = 485 mL

Now put all the given values in the above formula, we get:

0.0025=\frac{\text{Mass of }CaSO_4\times 1000}{136\times 485}

\text{Mass of }CaSO_4=0.16g

Thus, the mass in grams of calcium sulfate is 0.16 grams.

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A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off at and the t
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Therefore, the specific heat capacity of the iron is 0.567J/g.°C.

<em>Note: The question is incomplete. The complete question is given as follows:</em>

<em>A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm. </em>

<em> Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits</em>

Explanation:

Using the formula of heat, Q = mc∆T  

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c = specific heat capacity (J/g∙°C), ∆T = change in temperature (°C)

When the hot iron is placed in the water, the temperature of the iron and water attains equilibrium when the temperature stops changing at 27.6 °C. Since it is assumed that heat exchange occurs only between the iron metal and water; Heat lost by Iron = Heat gained by water

mass of iron  = 59.1 g, c = ?, Tinitial = 85.0 °C, Tfinal = 27.6 °C

∆T = 85.0 °C - 27.6 °C = 57.4 °C

mass of water = 100.0 g, c = 4.184 J/g∙°C, Tinitial = 23.0 °C, Tfinal = 27.6 °C

∆T = 27.6°C - 23.0°C = 4.6 °C

Substituting the values above in the equation; Heat lost by Iron = Heat gained by water

59.1 g * c * 57.4 °C  = 100.0 g * 4.184 J/g.°C * 4.6 °C

c = 0.567 J/g.°C

Therefore, the specific heat capacity of the iron is 0.567 J/g.°C.

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