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kramer
2 years ago
11

Question attached ! asap please

Chemistry
1 answer:
bezimeni [28]2 years ago
6 0

Answer:

A. copper is highly water soluble. It will turn into 5 different hydrates as it absorbs more and more water.

b. Glycerol is easily soluble in water, due to the ability of the polyol groups to form hydrogen bonds with water molecules

c. octane is considered to be non-polar, it will not be soluble in water, since water is a polar solvent. This will happen because octane (hydrocarbons in general) contains neither ionic groups, nor polar functional groups that can interact with water molecules.

d. Nitric acid decomposes into water, nitrogen dioxide, and oxygen, forming a brownish yellow solution.

e. Barium carbonate is a white powder. It is insoluble in water and soluble in most acids

Explanation:

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How are physical and chemical changes different
Wittaler [7]

Answer:

physic

Explanation:

physical chnages are chnages to the appearance

chemical chnages are chnages like

color temperature chnaging fromna gas to a liquid etc

3 0
3 years ago
Read 2 more answers
A sample of n2 effuses in 255 s. how long will the same size sample of cl2 take to effuse?
7nadin3 [17]
For this problem, we use Graham's Effusion Law to find out the rate of effusion of chlorine gas. The formula is as follows:

R₁/R₂ = √(M₂/M₁)

Let 1 be N₂ while 2 be Cl₂

255/R₂ = √(28/70.8)
Solving for R₂,
R₂ = 405.5 s

<em>Thus, it would take 405.5 s to effuse chlorine gas.</em>
4 0
2 years ago
What would be the result on the Insect population if the Bats and Mice were removed?
Ivanshal [37]

Answer:

We wouldn't have Coronavirus

Explanation:

5 0
2 years ago
Which of these equations is balanced, and which is not balanced? Explain how you can tell.
disa [49]

Answer:

a is balanced

Explanation:

a) reactant/ product

(2x2)=4c / 4c

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5 0
2 years ago
If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?
valkas [14]

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA     H⁺  +  A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid}     H⁺(aq)  +  A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ  + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

pH = 4.5199 = 4.52

Therefore the pH of the solution is 4.52

6 0
3 years ago
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