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Tatiana [17]
2 years ago
10

You start with 1 L of CO2 at standard temperature and pressure in a closed container. If you raise the temperature of the gas, t

he pressure will
Increase
Remain the same
Decrease
Chemistry
1 answer:
pychu [463]2 years ago
8 0

Answer:

Increase

Explanation:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

If the initial temperature and pressure is standard,

Pressure = 1 atm

Temperature = 273.15 K

then we increase the temperature to 400.0 K, The pressure will be,

1 atm / 273.15 K = P₂/400.0K

P₂ = 1 atm × 400.0 K / 273.15 K

P₂ = 400.0 atm. K /273.15 K

P₂ = 1.46 atm

Pressure is also increase from 1 atm to 1.46 atm.

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A substance that produces hydroxide ions when placed in water is a(n) _____.
bekas [8.4K]

Answer: A substance that produces hydroxide ions when placed in water is base.

Explanation:

Bases are the substance:

  • Which gives negatively charged hydroxide(OH^-) ions in aqueous solution.
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Where as acid gives positively charged hydronium ion(H^+) in aqueous solution.

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Why do nitrogen and oxygen form negative ions, not positive ones, in simple binary compounds.
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2 years ago
In an area in which electricity costs 8 cents/kilowatt-hour, a 5 kW clothes dryer runs for 90 minutes to dry a load of laundry.
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4 0
3 years ago
Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m
erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3}  =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2}  =0.06882molBr^-

Now, we compute the total moles of bromide:

n_{Br^-}=0.05456mol+0.06882mol\\\\n_{Br^-}=0.12338mol

Then, the total volume in liters:

150mL+175mL=325mL*\frac{1L}{1000mL} \\\\=0.325L

Therefore, the concentration of total bromide is:

M=\frac{0.12338mol}{0.325L}\\\\M=0.380M

Best regards!

8 0
3 years ago
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