Answer is: <span>The percent of KCl in the mixture is closest to </span><span>a) 40%.
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If we use 100 grams of mixture:
ω(K) = 44.20% ÷ 100% = 0.442.
m(K) = 0.442 · 100 g = 44.2 g.
n(K) = 44.2 g ÷ 39.1 g/mol.
n(K) = 1.13 mol.
n(KCl) + n(KNO₃) = n(K)
m(KCl) = x.
m(KNO₃) = y.
Two equtions:
1) x + y = 100 g.
2) m(KCl)/M(KCl) + m(KNO₃)/M(KNO₃) = 1.13 mol.
x/74.55 g/mol + y/101.1 g/mol = 1.13 mol.
From first equation find x = 100 - y and put in second equation:
(100 - y)/74.55 + y/101.1 = 1.13 /×101.1.
135.61 - 1.356y + y = 114.24.
0.356y = 22.37 g.
y = 62.83 g.
x = m(KCl) = 100 g - 62.83 g = 37.17 g.
Answer:
-219
Explanation:
1.5(339) - 1.5(485) = -219
Calcium ions have oxidation state 2+ => Ca (2+).
Bromime ions (bromide) have oxidation state 1- => Br (-).
So, to be neutral the compound has to have two Br (-) ions per each Ca(2+) ion.
That is represented in the chemical formula as Ca Br2, where the number 2 to the right of Br is a subscript meaning that there are two atoms of Br per each atom of Ca (the lack of subscript means 1 atom).
Answer: Ca Br2.
Answer:
67.88% is the percent yield.
Explanation:
Moles of methane = 
According to reaction,1 mole of methane gives 1 mole of carbon dioxide gas, then 3.75 moles of methane will give :
of carbon dioxide gas
Mass of 3.75 moles of carbon dioxide gas:
3.75 mol × 44 g/mol = 165 g
Theoretical yield of the carbon dioxide gas = 165 g
Experimental yield of the carbon dioxide gas = 112 g
The percentage yield of the reaction :
67.88% is the percent yield.
<span>You are given a cough syrup that contains 5.0% ethyl alcohol, c2h5oh, by mass and its density of the solution is 0.9928 g/ml. The molarity of the alcohol in the cough syrup is 21.55.</span>
