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shepuryov [24]
3 years ago
8

2. Calculate the molarity of a solution if there are 1.5 mol of NaCl in 2.3 L of solution.

Chemistry
1 answer:
rodikova [14]3 years ago
5 0

Answer:

0.65 M

Explanation:

Molarity (M) is the amount of moles of a substance per liter.

In other words, M = moles of substance / liters of solution

Since you're already given the moles of NaCl and the volume of the solution, you can plug those into the equation.

M = moles of NaCl / liters of solution

M = 1.5 mol NaCl / 2.3 L

M = 0.65 mol/L (or 0.65 M)

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A chemist designs a galvanic cell that uses these two half-reactions:
Virty [35]

Answer:

MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)

Explanation:

a)

MnO4⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn(aq)²⁺ + 4H2O (l)

b)

5Fe³⁺ (aq) +5e⁻ → 5Fe²⁺(aq)

c)

MnO4⁻ (aq) + 8H⁺ (aq) + 5Fe³⁺ (aq) →Mn(aq)²⁺ + 4H2O (l) + 5Fe²⁺(aq)

8 0
3 years ago
An ore is to be analyzed for its iron content by an oxidation-reduction titration with permanganate ion. A 4.230 g sample of the
Leya [2.2K]

Answer:

See explanation

Explanation:

The balanced redox reaction equation is;

8H+ + MnO4^- + 5Fe2+ ---------> Mn2+ + 5Fe3+ + 4H2O

Amount of KMnO4 reacted = 31.60/1000 * 0.05120 = 1.62 * 10^-3 moles

From the reaction equation;

1 mole of MnO4^- reacted with 5 moles of Fe2+

1.62 * 10^-3 moles will react with 1.62 * 10^-3 moles * 5/1 = 8.1 * 10^-3 moles

Mass of Fe2+ reacted = 8.1 * 10^-3 moles  *  56 g/mol

Mass of Fe2+ reacted = 0.45 g

Amount of iron in the sample =  0.45 g

Percentage of iron in the sample;

0.45 g/4.230 g  * 100 = 10.6 %

8 0
3 years ago
How many inches are in 2.0 miles
erastovalidia [21]
126,720 inches are in 2.0 miles
3 0
3 years ago
What Most Often Causes The Availability Of Water To Change?
Margaret [11]

Answer:the answer is C

Explanation:

8 0
3 years ago
A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid i
Alexeev081 [22]

Answer:

MM_{acid}=140.1g/mol

Explanation:

Hello,

In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:

n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}

Thus, solving for the moles of the acid, we obtain:

n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol

Then, by using the mass of the acid, we compute its molar mass:

MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol

Regards.

7 0
3 years ago
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