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3 years ago

Which one of these substances will exist in solution only as molecules?

Chemistry

1 answer:

lukranit [14]3 years ago

6 0

The substances which will exist in molecular form in solution are-

A. CH₃NH₂

D. CH₃OH

The methyl amine (CH₃NH₂) and methanol (CH₃OH) is organic compounds which are hardly become ionic in nature and prefers to form as molecular state in the solution phase.

On the other hand B. LiOH and C. NH₄OH are inorganic compound and highly ionic in nature and remain as ionic state in solution.

Those are-

LiOH (s) → Li⁺(s) + OH⁻ (s)

NH₄OH→ NH₄⁺ (s) + OH⁻ (s)

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Suppose the concentration of the standard stock solution is 50.04 ppm and that you delivered 4.60 mL of this standard stock solu

12345 [234]

Answer:

9.21 ppm

Explanation:

Considering

$concentration_{working\ solution}\times Volume_{working\ solution}=concentration_{stock\ solution}\times Volume_{stock\ solution}$

Given that:

$concentration_{working\ solution}=?$

$Volume_{working\ solution}=25.00mL$

$Volume_{stock\ solution}=4.60mL$

$concentration_{stock\ solution}=50.04 ppm$

So,

$concentration_{working\ solution}\times 25.00mL=50.04 ppm\times 4.60mL$

$concentration_{working\ solution}=\frac{50.04\times 4.60}{25.00}\ ppm=9.21\ ppm$

4 0

3 years ago

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our target reaction is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to flip reaction A.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$