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Dmitry [639]
3 years ago
15

At 0 Kelvin, gas particles do not possess any kinetic energy. TRUE FALSE

Chemistry
1 answer:
Maru [420]3 years ago
5 0
The answer is True :)
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Kidneys work with other body systems to maintain homeostasis by working with other body systems such as the respiratory system.
8 0
3 years ago
Each of the following values was read on an instrument of measuring device. In each case the last digit was estimated. Tell what
Drupady [299]

Answer:

<h3>160 cm</h3>

Explanation:

6 0
2 years ago
HEELLPP PLEASE IM BEGGING ILL GIVE BRAINLIEST
Arisa [49]

Answer:

C

Explanation:

You mix different thing together to make a new thing.

6 0
3 years ago
Will medal!!!
hjlf
Δ H reaction = q / n where q: amount of heat released and n is number of moles of substance.
q = m . C . ΔT where:
m = mass of substance (g)
C = Specific heat capacity (4.18)
ΔT = change in temperature = 24.25 - 23.16 = 1.09
q = 1000 x 4.18 x 1.09 = 4556 J = 4.556 kJ
number of moles (n) = Molarity (M) x Volume (L)
                                 = 0.185 M x 0.07 L = 0.01295 mole
Δ H = q / n = - (4.556 kJ / 0.01295 mole) = -351.8 kJ / mol
Note: it is exothermic reaction (-ve sign)  i.e. temperature is raised

5 0
3 years ago
Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products.
Alex_Xolod [135]

Answer:

a.

Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}

b.

Keq=[O_2]^3

c.

Keq=\frac{[H_3O^+][F^-]}{[HF]}

d.

Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}

Explanation:

Hello there!

In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:

a.

Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}

b.

Keq=[O_2]^3

c.

Keq=\frac{[H_3O^+][F^-]}{[HF]}

d.

Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}

Regards!

5 0
2 years ago
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