First, let's compute the number of moles in the system assuming ideal gas behavior.
PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles
At standard conditions, the standard molar volume is 22.4 L/mol. Thus,
Standard volume = 22.4 L/mol * 2.176 mol =<em> 48.74 L</em>
Answer: 0.0014 atm
Explanation:
Given that,
Original pressure of air (P1) = 1.08 atm
Original volume of air (T1) = 145mL
[Convert 145mL to liters
If 1000mL = 1l
145mL = 145/1000 = 0.145L]
New volume of air (V2) = 111L
New pressure of air (P2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
1.08 atm x 0.145L = P2 x 111L
0.1566 atm•L = 111L•P2
Divide both sides by 111L
0.1566 atm•L/111L = 111L•P2/111L
0.0014 atm = P2
Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm
I believe is different in pressure
That would be the first option Ca(OH)2 + H2SO4 → CaSO4 + 2H2O.
The Ca replaces the H2 in H2SO4, and the H2 replaces the Ca is Ca(OH)2.
Answer:
D is correct
Explanation:
because
we know that
density of lead is 11.36 g/cm3
and
density of tin is 7.31 g/cm3
so..
density of alloy by mixing 50/50
=(11.36+7.31)/2 g/cm3
=18.67/2 g/cm3
=9.33 g/cm3