Find the mass of 0.456 mol of Ca(OH)2 Please if you can show the work.
2 answers:
<h2>Answer:</h2>
<em>Hey, </em>
<h3><u>QUESTION)</u></h3>A) Let's first calculate the molar mass M of the molecule :
<em>M(Ca(OH)₂) = M(Ca) + 2M(O) + 2M(H) </em>
- M(Ca(OH)₂) = 40 + 2 x 16 + 2 x 1
- M(Ca(OH)₂) = 74 g/mol
B) Let's deduct the corresponding mass m :
<em>m = M x n </em>
- m = 74 x 0.456
- m = 33.744 g
- m ≈ 34,0 g
The mass of calcium hydroxide corresponding to this quantity of substance (in mol) is therefore 34 g.
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Answer:
1.346 v
Explanation:
1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:
(oxidation) →
E°=0.337 v
(reduction) →
E°=1.679 v
(overall) +8H^{+}_{(aq)}→
E°=1.342 v
2) Nernst Equation
Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:
Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.
The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give
E=1.346