The answer is: " 56 g CaCl₂ " .
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Explanation:
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2.0 M CaCl₂ = 2.0 mol CaCl₂ / L ;
Since: "M" = "Molarity" (measurement of concentration);
= moles of solute per L {"Liter"} of solution.
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Note the exact conversion: 1000 mL = 1 L .
Given: 250 mL ;
250 mL = ? L ? ;
250 mL * (1 L / 1000 L) = (250/1000) L = 0.25 L .
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(2.0 mol CaCl₂ / L ) * (0.25L) = (2.0) * (0.25) mol = 0.50 mol CaCl₂ ;
We have: 0.50 mol CaCl₂ ; Convert to "g" (grams):
→ 0.50 mol CaCl₂ .
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1 mol CaCl₂ = ? g ?
From the Periodic Table of Elements:
1 mol Ca = 40.08 g
1 mol Cl = <span>35.45 g .
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There are 2 atoms of Cl in " CaCl₂ " ;
→ Note the subscript, "2", in the " Cl₂ " ;
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So, to calculate the molar mass of "CaCl₂" :
40.08 g + 2(35.45 g) =
40.08 g + 70.90 g = 110.98 g ; round to 4 significant figures;
→ round to 111 g/mol .
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So:
→ 0.50 mol CaCl₂ = ? g CaCl₂ ? ;
→ 0.50 mol CaCl₂ * (111 g CaCl₂ / mol CaCl₂) ;
= (0.50) * (111 g) CaCl₂ ;
= 55.5 g CaCl₂ ;
→ round to 2 significant figures;
→ 56 g CaCl₂ .
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The answer is: " 56 g CaCl₂ " .
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Given in the problem is the mass of the liquid (500 grams) and the volume of the liquid (1000 ml = 1000 cm^3).
We can use these two givens to calculate the density of the liquid using the following rule:
density = mass / volume
density = 500 / 1000 = 0.5 grams / cm^3
Comparing the calculated density with the choices we have, we can deduce that the liquid is most likely to be propane with density 0.494 g / cm^3
Answer:
C3H7OH → C3H6 + H20
Explanation:
If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.
Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.
Answer:
Radioactive dating
Explanation:
Radioactive dating measures the ratio of C¹²(normal) and C¹⁴ (radioactive) in fossils to determine the exact age of the specimen. C¹⁴ is unstable and has a half life of 5730 years, so as the organism ages the amount of C¹⁴ diminishes but the amount of C¹² remains constant. The ratio of C¹⁴ to C¹² is used to analyze the decay of atomic particles to determine exact age.
The information given in the question is not enough to determine the acidity of the solution. This is because, acidity can only be found with the equation: pH = -log [H+].
In order to determine the acidity of the solution, the half titration point value is needed, this will make it possible to determine the value of H30+. If the half point titration value is known, then Ka will be equivalent to pH and the value will be evaluated using the equation: - log (1.6 * 10^-10).