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PIT_PIT [208]
3 years ago
7

Which of the following is a sign a chemical reaction has occurred? A liquid changes to a solid when cooled. A liquid forms bubbl

es when heated. Two clear liquids combine and mix together. Two clear liquids combine and turn cloudy.
Chemistry
2 answers:
faust18 [17]3 years ago
6 0

The last one “two clear liquid combine and turn cloudy” shows visual evidence that the two have combined and changed.

3241004551 [841]3 years ago
3 0

The best answer is, “Two clear liquids combine and turn cloudy.”

When water freezes form a liquid to a solid. That’s a physical change.

When water boils, it forms bubbles. That’s a physical change.

When you mix ethanol and water, they form a clear solution. That’s a physical change.

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Does bacteria need water to live
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No. bacteria only needs food to live along with other special needs depending on the type of bacteria
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4 years ago
Chemistry check please... Thank you!
I am Lyosha [343]
1) We apply the ideal gas equation:
PV = nRT
n = (21300 x 3/1000) / (8.314 x 323)
n = 0.024
Your answer is correct.

2) Total pressure = Partial pressure of Hydrogen + Partial pressure of water
134.7  = 122.4 + Pw
Pw = 12.3 kPa
Your answer is correct

3) The molar fraction, volume fraction and pressure fraction of gasses are the same thing.
Thus, percentage pressure of Oxygen = 10%
Pressure of Oxygen = 2.04 x 10⁴ x 0.1
= 2.04 x 10³ kPa
Your answer is correct

Well done!
3 0
3 years ago
PLEASE HELP ME ASAP!!!!
ipn [44]
The answer would be balanced
8 0
3 years ago
Read 2 more answers
Balance the following chemical equation by providing the correct coefficients HBr+Ba(OH)2 ———— BaBr2+H2O
stealth61 [152]

Answer:

2HBr + Ba(OH)2 = BaBr2 + 2H20

4 0
3 years ago
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
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