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Jet001 [13]
3 years ago
6

Which of the following is a common property of both strong acid and strong base?

Chemistry
2 answers:
kifflom [539]3 years ago
4 0

Answer:

They are good electrolytes

Gekata [30.6K]3 years ago
3 0

Answer:please can you help me with the last question that I asked I will help you if you help me please help

Explanation:

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What type of mixture is separated by effusion and condensation?
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Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when m
faust18 [17]

Answer:

56°

Explanation:

First calculate a:

a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}

The interplanar spacing can be calculated from:

d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}

The diffraction angle is determined from:

\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476

Solve for \theta

\theta=\sin ^{-1}(0.476)=28^{\circ}

The diffraction angle is:

2 \theta=2\left(28^{\circ}\right)=56^{\circ}

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3 years ago
Chlorine dioxide is used as a disinfectant and bleaching agent. In water, it reacts to form chloric acid (HClO3),
Sliva [168]
All done no worries

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What is the hardest mineral listed here? <br> A. quartz B. calcite C. gypsum D. corundum
a_sh-v [17]
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3 0
4 years ago
Read 2 more answers
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

4 0
3 years ago
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