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omeli [17]
3 years ago
6

Naomi has increased the pressure on a solution of liquid and gas in a closed container. What will this do to the gas in her solu

tion?
A. increase the amount

B. decrease the amount

C. increase the temperature

D. decrease the temperature
Chemistry
2 answers:
dsp733 years ago
6 0

Answer:

A.

Explanation:

dedylja [7]3 years ago
5 0

Answer:

increase the amount.

Hope this helps =)

Explanation:

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5. What happens to the mass when you change<br> the volume?<br> It decreases
anastassius [24]

Explanation:If the mass of the object stays the same but the volume of the object decreases then its density becomes greater. If the volume of the object stays the same but the mass of the object increases then its density becomes greater.

7 0
3 years ago
What evidence have u collected to explain the relationship between the Moon's revolution and lunar phrase?
Assoli18 [71]

Answer:

A lunar eclipse can only happen during a full moon.

Hope I helped :)

Explanation:

3 0
3 years ago
What is the lowest temperature a human can survive?
algol [13]
Suprysingly , Humans can survive weather well since we wear clothing.
A good estimate of how cold a human can survive would be in the range of -50 or lower.
HOPE IT HELPS!
3 0
3 years ago
Calculate the decrease in temperarure when 2.00 L at 20.0 degrees celsius is compressed to 1.00 L.
Serga [27]
I suspect that the pressure of this change is constant therefore

The equation is used from the combined gas law. (When pressure is constant both P's will cancel out P/P = 1)
V/T = V/T
Initial   Change

Initially we have 2L at 20 degress what temperature will be at 1L.

2/20 = 1/T
0.1 = 1/T
0.1T = 1
T = 1/0.1
T = 10 degress celsius.

Hope this helps if you won't be able to understand what is the combined gas law just tell me :).
3 0
3 years ago
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
3 years ago
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