Answer:
ΔH°c = -2219.9 kJ
Explanation:
Let's consider the combustion of propane.
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)
We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.
ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]
ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]
ΔH°c = -2219.9 kJ
Answer:
Th answer to your question is:
a) 3.5 x10⁻¹⁰ meters; 0.35 nm
b) 6857142.86 atoms
c) Volume = 2.06 x 10⁻²³ cm³
Explanation:
a) data
Uranium atoms = 3.5A°
meters
1 A° ---------------- 1 x 10 ⁻¹⁰ m
3.5A° --------------- x
x = 3.5(1 x10⁻¹⁰)/ 1 = 3.5 x10⁻¹⁰ meters
1 A° ------------------ 0.1 nm
3.5 A° ---------------- 0.35 nm
b) 2.4 mm
Divide 2,40 mm / uranium diameter
But, first convert 3,5A° to mm = 3.5 x 10⁻⁷ mm
# of uranium atoms = 2.4 / 3.5 x 10⁻⁷ = 6857142.86
c) volume in cubic cm
Convert 3.5A° to cm = 3.5 x 10⁻⁸
Volume = 4/3 πr³ = (4/3) (3.14)(1.7 x10⁻⁸)³
Volume = 2.06 x 10⁻²³ cm³
When there will be more quantity of co2 in air it will lead to more formation of carbonic acid and it will lower the Ph of rain 5.1 which leads to acid rain
Answer:
The correct answer is 28.2 %.
Explanation:
Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,
Ptotal = PO2 + PN2 + PHe
= 55 + 90 + 50
= 195 atm
The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100
Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,
%VO2 = (55/195) * 100 = 28.2%
%VN2 = (90/195) * 100 = 46%
%VHe = (50/195) * 100 = 25.64%
Hence, the percent oxygen by volume present in the blend is 28.2 %.
