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notsponge [240]
3 years ago
9

What is the final volume of NaOH solution prepared from 100.0 mL of 0.500 M NaOH if you wanted the final concentration to be 0.1

50 M?
Chemistry
1 answer:
suter [353]3 years ago
5 0

Answer:

The final volume of NaOH solution is 30ml

Explanation:

We all know that

              V1S1 = V2S2

             or V1= V2S2÷S1

             or  V1= V2×S2×1/S1

             or   V1=100×0.15×1/0.50

                    V1= 30

∴30 ml NaOH solution is required to prepare 0.15 M from 100ml 0.50 M NaOH solution.

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3 years ago
QUESTION 57
Tema [17]
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6 0
4 years ago
Read 2 more answers
Express 0.032 Ms in scientific notation using base units
bixtya [17]

0.032 Ms = 3.2 × 10^4 s

<em>Step 1</em>. Convert the number to scientific notation.

0.032 = 3.2 × 10^(-2)

<em>Step 2</em>. Convert the measurement to base units.

3.2 × 10^(-2) Ms × (10^6 s/1 Ms) = 3.2 × 10^4 s

6 0
3 years ago
Balance the following redox equations by the half-reaction method:
frozen [14]

Answer:

a)  Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

b) 2Bi(OH)3 + 3SnO2^2-  → 2Bi + 3SnO3^2- + 3H2O

c) Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

d) 2ClO3- + 2Cl- + 4H+   → Cl2 + 2H2O +2ClO2

e) 5 BiO3^- + 14 H+ + 2Mn^2+  → 5Bi^3+ + 7H2O + 2MnO4^-

Explanation:

<em>(a) Mn2 + H2O2 → MnO2 + H2O (in basic solution)</em>

Step 1: The half reactions

Oxidation: Mn2+ + 4OH- → MnO2 + 2H2O + 2e-

Reduction: H2O2 + 2e- + 2H2O  →  2H2O + 2OH-

Step 2: Sum of both half reactions

Mn2+ + 4OH- + H2O2  → MnO2 + 2H2O  + 2OH-

Step 3: the netto reaction

Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

<em>(b) Bi(OH)3 + SnO2^2-  → SnO3^2- + Bi (in basic solution)</em>

Step 1: The half reactions

Reduction:  Bi(OH)3 + 3e-  → Bi

Oxidation : Sno2^2-  → SnO3^2- +2e-

Step 2: Balance the half reactions

2* (Bi(OH)3 + 3e-  → Bi + 3OH-)

3* (Sno2^2- +2OH-  → SnO3^2- +2e- + H2O)

Step 3: The netto reaction

2Bi(OH)3 + 3SnO2^2-  → 2Bi + 3SnO3^2- + 3H2O

<em>(c) Cr2O7^2- + C2O4^2- → Cr^3+ + CO2 (in acidic solution)</em>

Step 1: The half reactions

Reduction: Cr2O7^2- + 6e-  → 2Cr+

Oxidation : C2O4^2- → 2CO2 + 2e-

Step 2: Balance the half reactions

Cr2O7^2- + 6e-  +14H+  → 2Cr+ +7H2O

3*(C2O4^2- → 2CO2 + 2e-)

Step 3: The netto reaction

Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

<em>(d) ClO3^- + Cl^− </em>→<em> Cl^2 + ClO^2 (in acidic solution)</em>

Step 1: The half reactions

Reduction: 2 ClO3^- + 10e- → Cl2

                      ClO3^- + e- → ClO2

 2 Cl- + 2ClO3^- +8e- →2Cl2

Oxidation: 2Cl- → Cl2 + 2e-

                   Cl- → ClO2 + 5e-

Cl- +ClO3^- → 2ClO2 + 4e-

Step 2: Balance the reactions

2Cl- + 2ClO3^- + 8e- + 12H+ → 2Cl2 + 6H2O

2* (Cl- + ClO3^- + H2O → 2ClO2 + 4e- + 2 H+)

Step 3: The netto reaction

2ClO3^- + 2Cl- + 4H+   → Cl2 + 2H2O +2ClO2

<em>(e) Mn^2 + BiO3^− </em>→<em> Bi^3 + MnO^4− (in acidic solution)</em>

Step 1: The half reactions

Reduction: BiO3^- + 2e- → Bi^3+

Oxidation : Mn^2+ → MnO4^- +5e-

Step 2: Balanced the reactions

5* ( BiO3^- + 2e- + 6H+ → Bi^3+ + 3H2O)

2* ( Mn^2+ + 4H2O →MnO4^- + 5e- + 8H+)

Step 3: The netto reaction

5 BiO3^- + 14 H+ + 2Mn^2+  → 5Bi^3+ + 7H2O + 2MnO4^-

6 0
4 years ago
What is the correct answer please
Anestetic [448]

Answer:

I think it's the second answer --If you increase the acidity..

I hope answer I can answer your question!

7 0
3 years ago
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