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nlexa [21]
3 years ago
11

If you decreased the volume of a sample of gas by a factor of three while maintaining a constant

Chemistry
2 answers:
anyanavicka [17]3 years ago
4 0

If you decreased the volume of a sample of gas by a factor of three while maintaining a constant pressure, how would the absolute temperature of the gas be affected?

✔ it would decrease threefold

A gas at 300 K and 4.0 atm is moved to a new location with a temperature of 250 K. The volume changes from 5.5 L to 2.0 L. What is the pressure of the gas at the new location?

✔ 9.2 atm

What is the partial pressure of 0.50 mol Ne gas combined with 1.20 mol Kr gas at a final pressure of 730 torr?

✔ 215 torr

Yuki888 [10]3 years ago
4 0

Answer:

I did the assignment

Explanation:

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How many gases are there that are not noble gases
natulia [17]

Answer:

Noble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og).

Explanation:

I pretty much covered it in my answer!

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8 0
3 years ago
The equation 2al(s) + 3br2(l) → 2albr3(s) is a(n) ______________ reaction.
Svetach [21]
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7 0
4 years ago
A cylinder is filled with 10.0L of gas and a piston is put into it. The initial pressure of the gas is measured to be 209.kPa. T
disa [49]

Answer : The final pressure of the gas will be, 26.8 kPa

Explanation :

According to the Boyle's law, the pressure of the gas is inversely proportional to the volume of the gas at constant temperature of the gas and the number of moles of gas.

P\propto \frac{1}{V}

or,

PV=k

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure of the gas = 209 kPa

P_2 = final pressure of the gas = ?

V_1 = initial volume of the gas = 10.0 L

V_2 = final volume of the gas = 78.0 L

Now put all the given values in this formula, we get the final pressure of the gas.

209kPa\times 10.0L=P_2\times 78.0L

P_2=26.8kPa

Therefore, the final pressure of the gas will be, 26.8 kPa

3 0
3 years ago
A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg
Brut [27]

Answer:  1.0\times 10^2g

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

K_f = freezing point constant = 3.96^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

Thus 1.0\times 10^2g urea was dissolved.

8 0
4 years ago
When rubbed against a silk cloth, a glass rod loses electrons. How does this affect the charges of the glass rod and the silk cl
AURORKA [14]

Answer:

Explanation:

The glass rod losses electrons because the silk cloth has a positive charge so it attracts the negative charge of the glass rod.

7 0
3 years ago
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