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OLga [1]
3 years ago
12

The boiling point of an aqueous solution is 101.54 ∘C. What is the freezing point

Chemistry
1 answer:
Hatshy [7]3 years ago
8 0

Answer:

101.54 .C.

Explanation:

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The following equation represents the type of reaction called
Mazyrski [523]

Answer:

Reduction

Explanation:

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Mn⁺⁷ +3e⁻    →    Mn⁴⁺

Mn gets three electrons , its oxidation state reduced from +7 to +4 so Mn gets reduced.

Examples:

Consider the following reactions.

4KI + 2CuCl₂  →   2CuI  + I₂  + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O   →  CO₂ + H₂

the oxidation state of carbon is +2 on reactant  side and on product side it becomes  +4 so carbon get oxidized.

H₂S + 2NaOH → Na₂S + 2H₂O

The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.

6 0
3 years ago
On a graph , which type of line shows a direct proportion?
svetoff [14.1K]
C - straight line
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5 0
3 years ago
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Define Acid dissociation constant constant
stiks02 [169]

Answer:

An acid dissociation constant, K a, (also known as acidity constant, or acid-ionization constant) is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for a chemical reaction ↽ − − ⇀ − + + known as dissociation in the context of acid–base reactions.

Explanation:

6 0
3 years ago
Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to air
Yanka [14]

Answer: a. 79.6 s

b. 44.3 s

c. 191 s

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{8.7\times 10^{-3}s^{-1}}=79.6s

b) for completion of 32% of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-32}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{68}

t=44.3s

c) for completion of 81 % of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-81}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{19}

t=191s

4 0
3 years ago
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a specific hypothesis to test with a controlled experiment

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