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Brums [2.3K]
3 years ago
5

2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.

Chemistry
1 answer:
AysviL [449]3 years ago
7 0

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

(b) Moles of H_2SO_4 can be calculated as:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For H_2SO_4 :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of H_2SO_4 :

Moles=0.1450 \times {10\times 10^{-3}}\ moles

Moles of H_2SO_4  = 0.00145 moles

From the reaction,

1 mole of H_2SO_4 react with 2 moles of NaOH

0.00145 mole of H_2SO_4 react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 /  21.37×10⁻³  M = 0.1357 M

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An element has an atomic number of 27. How many protons and electrons are in a neutral atom of the element
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27 and 32

Explanation:

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6.10 g of a solid chemical reacts with a 10 g of a liquid chemical, the
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3 years ago
If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?
Nata [24]

Answer:

The correct answer is option B.

Explanation:

3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2

Moles of NBr_3 = 40 mol

Moles of NaOH = 48 mol

According to reaction, 3 moles of NaOH reacts with 2 moles NBr_3

Then ,48 moles of NaOH will reacts with:

\frac{2}{3}\times 48 mol=32 mol of NBr_3

Then ,40 moles of NaBr_3 will reacts with:

\frac{3}{2}\times 40 mol=60 mol of NaOH

As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.

Moles left after reaction = 40 mol - 32 mol = 8 mol

Hence, the NBr_3 is an excessive reagent.

6 0
3 years ago
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The pOH of an aqueous solution at 25°C was found to be 1.20. The pH of this solution is . The hydronium ion concentration is M.
frosja888 [35]

Answer:

pH = 12.80

[H3O+] = 1.58 * 10^-13 M

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Explanation:

Step 1: Data given

pOH = 1.20

Temperature = 25.0 °C

Step 2: Calulate pH

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.20 = 12.80

Step 3: Calculate hydronium ion concentration

pH = -log[H+] = -log[H3O+]

12.80 = -log[H3O+]

10^-12.80 = [H3O+] = 1.58 * 10^-13 M

Step 4: Calculate the hydroxide ion concentration

pOH = 1.20 = -log [OH-]

10^-1.20 = [OH-] = 0.063M

Step 5: Control [H3O+] and [OH-]

[H3O+]*[OH-] = 1* 10^-14

1.58 *10^-13 * 0.063 = 1* 10^-14

5 0
3 years ago
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