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This is the balanced equation.
<span>2 C4H10g + 13 O2g ---> 8 CO2g + 10 H2Og</span>
D) The broadest group of organization is known as the domain.
MA= output force/input force
MA= 100N/20N
MA= 50
Answer:
14.434 r.a.m.
Explanation:
- The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.
∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)
atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.
atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.
∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.
Answer:
V₂ = 1.92 L
Explanation:
Given data:
Initial volume = 0.500 L
Initial pressure =2911 mmHg (2911/760 = 3.83 atm)
Initial temperature = 0 °C (0 +273 = 273 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
by putting values,
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm
V₂ = 522.795 atm .L. K / 273 K.atm
V₂ = 1.92 L
