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Ann [662]
3 years ago
14

What volume of 0.0748 M perchloric acid can be neutralized with 115 mL of 0.244 M sodium hydroxide?

Chemistry
1 answer:
Irina18 [472]3 years ago
5 0

Answer:

0.375 L

Explanation:

We know that at neutralization, the number of mol  of acid must equal the number of equivalents of base.

This is a reaction 1:1 acid to base:

HClO₄ + NaOH ⇒ NaClO₄ + H₂O

We re given the  moles  of the base indirectly since we know the volume and molarity. From there we can calculate the volume of HClO₄.

Moles NaOH = 0.115 L x 0.244 M = 0.115 L x 0.244 mol/L =0.028 mol

Thus we require 0.028 mol of HClO₄ in the pechloric acid solution:

Molarity = # moles / V ⇒ V = # moles / M

V = 0.028 mol / 0.0748 mol/L = 0.375 L

Note  that this problem can be solved in just one step since

M(HClO₄) x V(HClO₄) = M(NaOH) x V(NaOH)  ⇒

V(HClO₄) = M(NaOH) x V(NaOH) / M(HClO₄)

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1.5M NaOH so we've 1.5 moles of NaOH in 1L of solution

1L = 1000 ml

1.5 moles of NaOH ------------in------------- 1000 ml
0.75 moles of NaOH ----------in---------------x ml
x = 500 ml

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5 0
3 years ago
For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K i
frosja888 [35]

Answer:

Explanation:

The formula of the reaction:

            KClO₂ → KCl + O₂

To assign oxidation numbers, we have to obey some rules:

  1. Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
  2. The charge on simple ions signifies their oxidation number.
  3. The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.

The oxidation number of K in KClO₂:

                                   K + (-1) + 2(-2) = 0

                                    K-5 = 0

                                    K = +5

The oxidation number of K in KCl:

                                K + (-1) = 0

                                K = +1

The oxidation number Cl in KClO₂ is -1

For Cl in KCl, the oxidation number is -1

For O in KClO₂, the oxidation number is (2 x -2) = -4

For O in O₂, the oxidation number is 0

K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.

O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.

7 0
3 years ago
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How much energy is released when 0.40 mol C6H6(g) completely reacts with oxygen.
Mashutka [201]
You have to find the gram formula mass of C6H6 then do mass (g) = mol x GFM
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3 years ago
What is the temperature of liquid nitrogen?
motikmotik
Really darn cold lol
6 0
3 years ago
How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction
nekit [7.7K]

Answer : The volume of NaOH required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

Hence, the volume of NaOH required to neutralize is, 340 mL

4 0
3 years ago
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