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madreJ [45]
4 years ago
14

Upon heating a copper sample will expand, leading to a lower density? A) TrueB) False

Chemistry
1 answer:
Leona [35]4 years ago
8 0

Answer;

The above statement is true

upon heating a copper sample will expand, leading to a lower density

Explanation;

-The density of solids decreased with increase in temperature and vice versa. The increase in temperature causes the volume of the solid to increase which as a result decreases the density as Density=Mass/Volume. The temperature of a body is the average kinetic energy of the molecules present in it.

In other words; The temperature of a body is the average kinetic energy of the molecules present in it. Therefore; when heat is supplied ( or temperature is increased) the average kinetic energy increases which increases the volume and thus density decreases.

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Which of the following reactions will not take place?
lana [24]

Answer:

Explanation:

d because its already a liquid as well as the acid

5 0
4 years ago
Classify these salts as acidic, basic, or neutral. kcl, nh4br, k2co3, nacn, liclo4? this is what i put but got it wrong: nh4br =
olchik [2.2K]
I'll just list down the acidic, basic, and neutral salts.

ACIDIC
<span>THE CATION: is the conjugate acid of a weak base
THE ANION: Conjugate Base of A Strong Acid

</span>Ammonium Sulfate - <span>(NH₄)₂SO₄
</span>Ammonium Chloride - NH₄Cl
BASIC
<span>THE CATION: comes from the cation of a strong Base (i.e. Na+, K+, Li+)
The Anion: comes from the conjugate base of a weak acid 

</span>Sodium Acetate - CH₃COONaSodium Phosphate - <span>Na₃PO₄
</span>Calcium Acetate - (CH₃COO)₂Ca
NEUTRAL
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THE ANION: comes from a strong acid

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5 0
4 years ago
Which of these is a property of acid?
Taya2010 [7]
C. They Typically Cantain AN -Oh group
3 0
4 years ago
Which of the following is not a characteristic of
emmainna [20.7K]

Answer:

high density

Explanation:

5 0
4 years ago
The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process
denis23 [38]

Answer:

0.1066 hours

Explanation:

A common pesticide degrades in a first-order process with a rate constant (k) of 6.5 1/hours. We can calculate its half-life (t1/2), that is, the times that it takes for its concentration to be halved, using the following expression.

t1/2 = ln2/k

t1/2 = ln2/6.5 h⁻¹

t1/2 = 0.1066 h

The half-life of the pesticide is 0.1066 hours.

7 0
4 years ago
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