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Allisa [31]
3 years ago
5

A sample of 0.800 mol of nitrogen has a pressure of 0.600 atm and a

Chemistry
1 answer:
schepotkina [342]3 years ago
5 0

Answer:

274K

Explanation:

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Potassium + water = potassium hydroxide + hydrogen​
Paul [167]

i hope it help you a lot tell it is correct or not

7 0
2 years ago
A tank of oxygen has a volume of 1650 L. The temperature of the gas inside is 35?C. If there are 9750 moles of oxygen in the tan
Paul [167]

Answer:

2192.64 PSI.

Explanation:

  • From the general law of ideal gases:

<em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the container in L (V = 1650 L).

n is the no. of moles of the gas in mol (n = 9750 mol).

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature of the gas in (T = 35°C + 273 = 308 K).

∴ P = nRT/V = (9750 mol)(0.082 L.atm/mol.K)(308 K)/(1650 L) = 149.2 atm.

  • <u><em>To convert from atm to PSI:</em></u>

1 atm = 14.696 PSI.

<em>∴ P = 149.2 atm x (14.696 PSI/1.0 atm) = 2192.64 PSI.</em>

4 0
3 years ago
Howtocalculatethevolumeofcarbondioxideproducedwhen400gofmarblewereats.t.p<br>​
lys-0071 [83]

Answer:

so 0.15 moles X 22.4 dm3/mole=3.36 dm3. Next we find the moles of hexane combusted, and then the moles of CO2. Finally, we find the volume of CO2 using the fact that at STP, 1 mole of gas = 22.4 dm3.

3 0
3 years ago
Hshshsbxbxbbxbzhzhxbdidu
Marysya12 [62]
Rffffffffffffffffffffffffffffffrrrrrrrrrrrrrrrrrrrrrrfffffffffffffffffffrrrrrrrrrrrrrrrrr
7 0
3 years ago
Read 2 more answers
A solution is made by dissolving 0.656 mol of nonelectrolyte solute in 869 g of benzene. calculate the freezing point, tf, and b
solmaris [256]
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.

1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.

2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
</span>
3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>

</span>
4 0
3 years ago
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