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Nady [450]
3 years ago
7

Examples of a pure substances

Chemistry
2 answers:
KonstantinChe [14]3 years ago
8 0

Answer:

Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements

Explanation:

DiKsa [7]3 years ago
3 0

Answer:

Explanation:

Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements.

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The substances present before a chemical reaction takes place are called?
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What is the definition of an atmosphere? a portion of the electromagnetic spectrum that makes the sky look blue the blanket of g
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the blanket of gases that surrounds Earth and some other planets

Explanation:

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3 years ago
Lowering an object decrease is potential​
Tresset [83]

Answer:

Lowering the object near the ground decreases its <u>potential energy.</u>

<u></u>

Explanation:

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Change in shape and size :  When  you compress the spring , potential energy is introduced in it . So it expand quickly when you remove your hand.

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Potential energy(P) is given by the formula :

P = mgh

where ,

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8 0
3 years ago
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.
yaroslaw [1]

Answer:

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

Explanation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

initially

3.0 atm      0              0

At equilibrium

(3.0-2p)     p               3p

Equilibrium partial pressure of nitrogen gas = p = 0.90 atm

The expression of a pressure equilibrium constant will be given by :

K_p=\frac{p_{N_2}\times (p_{H_2})^3}{(p_{NH_3})^2}

K_p=\frac{p\times (3p)^3}{(3.0-2p)^2}

=\frac{0.90 atm\times (3\times 0.90 atm)^3}{(3.0-2\times 0.90 atm)^2}

K_p=12.30\approx 12.

12. is the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.

3 0
3 years ago
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