Answer:
49.35 mL
Explanation:
Given: 56.2 mL of gas
To find: volume that 56.2 mL of gas at 820 mm of Hg would occupy at 720 mm of Hg
Solution:
At 820 mm of Hg, volume of gas is 56.2 mL
At 1 mm of Hg, volume of gas is 
At 720 mm of Hg, volume of gas is 
Answer:
A thermochemical equation for the combustion of propane (C3H8)(C3H8) is written as follows:
C3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔH∘rxnC3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔHrxn∘ = -2202.0 kJ/mol
The value given for ΔH∘rxnΔHrxn∘ means that:
a. the reaction of one mole of propane absorbs 2202 kJ of energy from the surroundings.
b. the reaction is endothermic.
c. the enthalpy of formation of propane is 2202 kJ/mol.
d. the reaction of one mole of propane releases 2202 kJ of energy to the surroundings.
e. None of these.
First, we have to calculate the number of moles of H2SO4 in the solution:
V=60 mL = 0.06 L
c=5.85 mol/L
n=V×c=0.06×5.85=0.351 mol
Then we need to find the molar mass of H2SO4:
2×Ar(H) + Ar(S) + 4×Ar(O) =
=2 + 32 + 64 = 98 g/mol
Finally, we need to find the mass of H2SO4:
m=0.351 × 98 = 34.398 g
<span>PV=nRT= a universal constant
For any condition
P1V1/n1T1=R
and
P2V2/n2T2=R
i.e
P1V1/n1T1=P2V2/n2T2
Becomes
V1/n1=V2/n2
rearranging and solving
V2=V1X(n2/n1)= 750 mLx((0.65+0.35)/(0.65))=1200ml=1.2L...2 sig figs</span>
Answer:
first one-liquids dont have a fixed shape and solids do
