Answer:
The concentration of the chloride ion is 1.8 mM
Explanation:
Step 1: Data given
An aqueous solution contains 3.6 mM of total ions.
Step 2: The equation
NaCl(aq) → Na+(aq) + Cl-(aq) : two ions
Step 3: Calculate the concentration of the chloride ion
Two ions = 3.6 mM
Chloride ion = 3.6/2 mM= 1.8 mM
Sodium ion = 3.6/2 mM = 1.8 mM
The concentration of the sodium ions is 1.8 mM
The concentration of the chloride ion is 1.8 mM
The total concentration = 3.6 mM
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Substance c will have the lowest temperature
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.
