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4 years ago

Which acid has the greatest acid dissociation constant?

Chemistry

2 answers:

Kobotan [32]4 years ago

6 0

Nitric acid has the more dissociation constant

Dmitriy789 [7]4 years ago

5 0

Answer:

nitric acid

Explanation:

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at 30°C 80 g of sodium nitrate is dissolved in 100 g of water is a solution saturated and unsaturated or supersaturated

soldi70 [24.7K]

Answer:

Explanation:

thats the answer

4 0

4 years ago

In the alkaline phosphatase assay, you will you be monitoring the formation of product by measuring the absorbance of the soluti

blondinia [14]

Answer:

a. At pH 9, the product, p-nitrophenol, will be ionized, the solution will appear yellow in color, and thus can be monitored at the wavelength of maximum absorption for the phenolate ion which is 400nm

Explanation:

In alkaline phosphatase assay, the hydrolysis of p nitrophenyl phosphate to p nitrophenol happens. When the ph is 9, the product which is p nitrophenol would undergo ionization. The solution is going to appear to be of yellow and it can be monitored at a wavelength for maximum absorption of phenolate ions at 400nm.

Option A is the answer to the question.

5 0

3 years ago

Stoichiometry is best defined as the

Oxana [17]

Stoichiometry is “quantitative relationship” among the “reactants” and the “products” in a “chemical reaction”.

<u>Explanation</u>:

In stoichiometry “stoicheion” means element and “metron” means measure in Greek. The stoichiometric calculation depends upon “stoichiometric coefficients” in a “chemical equation” which can be explained as the “number of moles” of each substance (reactants or products). Stoichiometric calculation is done as follows:

For example reaction between nitrogen and hydrogen to form ammonia as

$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})$

Here stoichiometric coefficients show that “one molecule of nitrogen” reacts with “three molecules of hydrogen” to form “two molecules of ammonia”. Multiplying Avogadro number $6.022 \times 10^{23}$ to no of molecules in equation:

$\text { For } \mathrm{N}_{2}: 1 \times 6.022 \times 10^{23}=1 \mathrm{mol}$

$\text { For } 3 \mathrm{H}_{2}: 3 \times 6.022 \times 10^{23}=3 \mathrm{mol}$

$\text { For } 2 \mathrm{NH}_{3}: 2 \times 6.022 \times 10^{23}=2 \mathrm{mol}$

Taking molar masses into consideration:

$\text { For } \mathrm{N}_{2}: 1 \times 28.0=28.0 \mathrm{gm}$

$\text { For } 3 \mathrm{H}_{2}: 3 \times 2=6.0 \mathrm{gm}$

$\text { For } 2 \mathrm{NH}_{3}: 2 \times 17=34 \mathrm{gm}$

Hence balanced equation gives stoichiometric coefficients which gives proportion by moles.

8 0

3 years ago

Silicon has a relatively large latent heat of vaporization of 12800 J/g. This indicates that silicon has:

beks73 [17]

I believe it means that it takes a lot of heat in joules to make silicon vaporize so it stays solid until a great deal of heat has been added. By comparison, water has a latent heat of about 2260 joules per gram so vaporizes much more readily than silicon.

3 0

3 years ago

RATE LAW QUESTION !

vivado [14]

In general, we have this rate law express.:

$\mathrm{Rate} = k \cdot [A]^x [B]^y$
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation $\Delta \mathrm{Rate} = \Delta [B]^y$

$3.09 = (3)^y \implies y \approx 1$

To this point, $\mathrm{Rate} = k \cdot [A]^x [B]^1$

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for $\Delta \mathrm{Rate} = \Delta [A]^x$

$4= (2)^x \implies x = 2$

the rate law is

Rate = k·[A]²[B]

6 0

3 years ago