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notsponge [240]
3 years ago
14

Any change that alters a substance without changing it into another substance.

Chemistry
1 answer:
Stolb23 [73]3 years ago
5 0
I think its physical change hope this helps
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How does the addition of salt to solid ice affect the melting transition from solid to liquid?
serious [3.7K]

Answer:

Im sure the answer is c

Hope this helped have a good day :) please mark me the brainliest answer?

7 0
3 years ago
Read 2 more answers
The amount of energy that is transferred when one mole of a compound is formed from its component elements in their standard sta
amm1812
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5 0
3 years ago
You have a graduated cylinder with 10 mL of water in it.
Law Incorporation [45]

Answer:

<h3>The answer is 10 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 300 g

volume = final volume of water - initial volume of water

volume = 40 - 10 = 30 mL

We have

density =  \frac{300}{30}  =    \frac{30}{3}  \\

We have the final answer as

<h3>10 g/mL</h3>

Hope this helps you

5 0
3 years ago
A manganese electrode was oxidized electrically. If the mass of the electrode decreased by 225 mg during the passage of 1580 cou
qwelly [4]

Answer:

<u>Oxidation state of Mn = +4</u>

Explanation:

Atomic mass of Mn = 55g/mol

From Faraday's law of electrolysis,

Electrochemical equivalent = \frac{mass}{charge}

i.e Z = \frac{m}{Q} = \frac{0.225g}{1580C} = 0.0001424 g/C

But Equivalent weight, E = atomic mass ÷ valency  = Z × 96,485

⇒ \frac{55}{valency} = 0.0001424 × 96,485

<u>∴ Valency of Mn = +4</u>

8 0
3 years ago
(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2
maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ =  (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

5 0
3 years ago
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