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AVprozaik [17]
3 years ago
15

In the laboratory you dissolve 13.9 g of potassium phosphatein a volumetric flask and add water to a total volume of 250mL.

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

Molarity of the solution? 0,262 M.

Concentration of the potassium cation? 0,786 M.

Concentration of the phosphate anion? 0,262 M.

Explanation:

Potassium phosphate (K₃PO₄; 212,27 g/mol) dissolves in water thus:

K₃PO₄ → 3 K⁺ + PO₄³⁻ <em>(1)</em>

Molarity is an unit of chemical concentration given in moles of solute (K₃PO₄) per liters of solution.

There are 250 mL of solution≡0,25 L

The moles of K₃PO₄ are:

13,9 g of K₃PO₄ ×\frac{1mol}{212,27 g} = 0,0655 moles of K₃PO₄

The molarity of the solution is:

\frac{0,0655 moles}{0,25L} = 0,262 M

In (1) you can see that 1 mole of K₃PO₄ produces 3 moles of potassium cation. The moles of potassium cation are:

0,0655 moles×3 = 0,1965 moles

The concentration is:

\frac{0,1965 moles}{0,25L} = 0,786 M

The moles of K₃PO₄ are the same than moles of PO₄³⁻, thus, concentration of phosphate anion is the same than concentration of K₃PO₄. 0,262 M

I hope it helps!

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zysi [14]
            <span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)

grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required

Since there are only 45.0 grams HCl, then HCl is the limiting reactant.

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7 0
3 years ago
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0.352 g sample of a diprotic acid is dissolved in water and titrated with 0.150 M NaOH.0.150 M NaOH. What is the molar mass of t
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<h3>Answer:</h3>

128.94 g/mol

<h3>Explanation:</h3>

Given;

Mass of diprotic acid, H₂X =0.352 g

Molarity of NaOH = 0.150 M

Volume of the NaOH = 36.4 mL

We are required to calculate the molar mass of the acid.

Note: A diprotic acid is an acid that contains 2 replaceable hydrogen atoms

<h3>Step 1: Write the balanced equation fro the reaction;</h3>

The balanced equation for the reaction between the diprotic acid and NaOH will be;

H₂X(aq) + 2NaOH(aq) → Na₂X(aq) + 2H₂O(l)

<h3>Step 2: Determine the number of moles of NaOH used </h3>

Given the molarity and volume of NaOH we can calculate the number of moles;

Moles of NaOH = Molarity × Volume

                          = 0.150 M × 0.0364 L

                         = 0.00546 moles

<h3>Step 3: Use the mole ratio to determine moles of the acid </h3>

From the equation;

1 mole of the acid reacts with 2 moles of NaOH

Therefore; H₂x : NaOH = 1 : 2

Moles of H₂x = 0.00546 moles ÷ 2

                      = 0.00273 moles

<h3>Step 4: Determine the molar mass of the acid.</h3>

Molar mass is the mass equivalent to 1 mole of a compound or element

From our calculations;

0.00273 moles = 0.352 g of the acid;

Therefore, mass in 1 mole ;

= 0.352 g ÷ 0.00273 moles

= 128.94 g/mol

Thus, the molar mass of the diprotic, H₂X is 128.94 g/mol

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3 years ago
In the laboratory, a student dilutes 13.5 mL of a 11.6 M hydroiodic acid solution to a total volume of 100.0 mL. What is the con
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Answer: The concentration of the diluted solution is 1.566M.

Explanation:

The dilution equation is presented as this: M_{s} V_{s} =M_{d} V_{d}.

·M= molarity (labeled as M)

·V= volume (labeled as L)

·s= stock solution (what you started with)

·d= diluted solution (what you have after)

Now that we know what each part of the formula symbolizes, we can plug in our data.

11.6M*13.5mL=M_{d} *100mL

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Now that we have the conversions, let's plug them into the equation.

11.6M*0.0135L=M_{d} *0.1L

The only thing that we need to do now is actually solving the answer.

M_{d} =\frac{11.6M*0.0135L}{0.1L}       M_{d} =1.566M

From the work shown above, the answer is 1.566M.

I hope this helps!! Pls mark brainliest :)

7 0
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d) Z = 20, A = 42

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Answer: 8.830418848725065

Explanation:

8.830418848725065

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