830 mL
The volume of an 2.3 m solution with 212 grams of calcium chloride (cacl2) dissolved is 830 mL.
The solution has a concentration of 2.3 mol/L.
<h3>a) Moles of CaCl2</h3>
Molar mass of CaCl2 = 110.98 g/mol
Moles of CaCl2 = 212 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2)
= 1.910 mol CaCl2
<h3>b) Volume of solution</h3>
V = 1.910 mol CaCl2 x (1 L solution/2.3 mol CaCl2) = 0.83 L solution
= 830 mL solution
<h3>How much CaCl2 is there in the solution by molarity?</h3>
- The number of moles is 0.125 x 2 = 0.25 mol since the molarity is 2.0M.
- To get the answer of 27.745 g, simply multiply this by the molar mass of calcium chloride, which is 110.98 g/mol.
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<h2>The required option d) "specific heat" is correct.</h2>
Explanation:
- To raise the temperature of any substance or material of certain mass to respective temperature it requires some amount of heat.
- Specific heat is the amount of heat necessary to raise the temperature of the substance of 1 gram to 1 Kelvin.
- It is the amount of heat which is required to raise the temperature per unit mass to per unit temperature.
- Thus, the required "option d) specific heat" is correct.
<u>Answer:</u> The moles of oxygen and carbon dioxide in air is
and
respectively
<u>Explanation:</u>
To calculate the number of moles, we use the equation:

Given mass of atmosphere = 
Average molar mass of atmosphere = 28.96 g/mol
Putting values in above equation, we get:

We know that:
Percent of oxygen in air = 21 %
Percent of carbon dioxide in air = 0.0415 %
Moles of oxygen in air = 
Moles of carbon dioxide in air = 
Hence, the moles of oxygen and carbon dioxide in air is
and
respectively
The answer would be A be cause it is the most reasonable answer