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Allisa [31]
3 years ago
11

which law states that the volume and temperature are inversly proportional in a fixed quantity of gas

Chemistry
1 answer:
svlad2 [7]3 years ago
7 0
The Charles Law : The Temperature-Volume Law

(One of the Charles laws is the temperature volume law)




Hope this helped

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When a solution of AgNO3 is mixed with a solution of NaBr (multiple options could be correct):
alex41 [277]

Answer:

After a few minutes pass, the concentration of Ag+ and Br- will be lower than when the two solutions were first mixed.

AgBr precipitate will spontaneously form.

Explanation:

After the net ionic equation AgBr forms

7 0
3 years ago
A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide. What is the partial pressure of oxygen
Kay [80]
Answer is: pressure of oxygen is 31,3 kPa.
The total pressure<span> of an ideal gas mixture is the sum of the </span>partial pressures<span> of the gases in the mixture.
p(mixture) = p(helium) + p(oxygen) + p(carbon dioxide).
p(oxygen) = p(mixture) - (p(helium) + p(carbon dioxide)).
p(oxygen) = 101,4 kPa - (68,7 kPa + 1,4 kPa).
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p(oxygen) = 31,3 kPa.

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3 0
3 years ago
A radioactive substance of mass 768g has a half life of 3 years. After how many years does it leave only 6g undecayed?​
vagabundo [1.1K]

Answer:

The answer is 21 years.

Explanation:

7×3

8 0
3 years ago
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The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
3 years ago
Which describe beta decay? Check all that apply 1.In beta decay, a proton becomes a neutron and a positron 2.In beta decay, a ne
ivanzaharov [21]

haha thanks again for the help.

3 0
3 years ago
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