The ostrich runs 21 N then goes back by 25 S
The displacement is -4 S
21 - 25 = -4
Answer:
V = 42.6 L
Explanation:
Given data:
Number of moles of Cl₂ = 1.9 mol
Temperature and pressure = standard
Volume occupy = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm × V = 1.9 mol ×0.0821 atm.L /mol.K × 273.15 k
V = 42.6 atm.L / 1 atm
V = 42.6 L
Answer is: 12,6% (1/8) <span>percentage of the sample will remain.
</span>c₀ - initial amount of C-14.<span>
c - amount of C-14 remaining
at time.
t = 5700</span> y.<span>
First calculate the radioactive decay rate constant λ:
λ = 0,693 ÷ t = 0,693 ÷ 5700</span> y = 0,000121 1/y = 1,21·10⁻⁴ y.
c = c₀·e∧-λ·t.
c = 2000 · e∧-(0,000121 1/y · 17100 y).
c = 252 g.
ω = 252 g ÷ 2000 g = 0,126 = 12,6%.
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</span>
Answer:
6ΔG°(f) H₂O = -229 Kj/mol
Explanation:
4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)
ΔG°(f) 4mol(-16.66Kj/mol) | 5mol(0Kj/mol) || 4mol(+86.71Kj/mol) | 6ΔG°(f) H₂O
Hess's Law
ΔG°(Rxn) = ∑ΔG°(f) Products - ∑ΔG°(f) Reactants
-957.9 Kj = [(4mol(+86.71Kj/mol)) + 6ΔG°(f) H₂O(g)] - [4mol(-16.66Kj/mol) + 5mol(0Kj/mol)]
-957.9 Kj = [4(86.7)Kj + 6ΔG°(f) H₂O] - [4(-16.66)Kj] = 346.84Kj + 6ΔG°(f) H₂O + 66.64Kj
ΔG°(f) H₂O = ((-957.9 - 346.84 -66.64)/6)Kj = -228.56 Kj ≅ -228.6 Kj*
*Verified with Standard Heat of Formation Table
