I suppose it would be forest because in order to have organic matter the soil needs to be rich and fertile,therefore it is forest.
Answer:
See Explanation
Explanation:
Hence the mass defect is;
[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]
= 236.0526 - 235.85361
= 0.19899 amu
Since 1 amu = 1.66 * 10^-27 Kg
0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg
Binding energy = Δmc^2
Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J
ii) 
Hence the mass defect is;
[10.01294 + 1.00867] - [7.01600 + 4.00260]
= 11.02161 - 11.0186
= 0.00301 amu
Since 1 amu = 1.66 * 10^-27 Kg
0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg
Binding energy = Δmc^2
Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J
Answer:
The equilibrium constant for the reversible reaction = 0.0164
Explanation:
At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.
The reaction is given as
A ⇌ B
Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]
The rate of forward reaction = |r₁| = k₁ [A]
The rate of backward reaction = |r₂| = k₂ [B]
(Taking only the magnitudes)
where k₁ and k₂ are the forward and backward rate constants respectively.
k₁ = 0.010 s⁻¹
k₂ = 0.0610 s⁻¹
|r₁| = 0.010 [A]
|r₂| = 0.016 [B]
At equilibrium, the rate of forward and backward reactions are equal
|r₁| = |r₂|
k₁ [A] = k₂ [B] (eqn 1)
Note that equilibrium constant, K, is given as
K = [B]/[A]
So, from eqn 1
k₁ [A] = k₂ [B]
[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164
K = [B]/[A] = (k₁/k₂) = 0.0164
Hope this Helps!!!
Note that this is occurring at STP, where 22.4L of any gas is equal to 1mol of that gas.
First, convert the liters of O₂ to moles of O₂ using the conversion factor 22.4LO₂ = 1molO₂.
8.6LO₂ × 1molO₂/22.4LO₂
= 8.6/22.4
≈ 0.3839molO₂
Next, convert moles of O₂ to moles of H₂O. In the balanced equation, the coefficients show that there are 2 moles of H₂O for every mole of O₂. So, use the conversion factor 1molO₂ = 2molH₂O.
0.3839molO₂ × 2molH₂O/1molO₂
= 0.3839 × 2
= 0.7678molH₂O
Finally, convert the moles of H₂O to liters of H₂O using the same conversion factor from before, 22.4LH₂O = 1molH₂O.
0.7678molH₂O × 22.4LH₂O/1molH₂O
= 0.7678 × 22.4
≈ 17LH₂O
So, the answer is 17 liters of gaseous water is collected! Note that its rounded to 17 because the measurement given in the problem has 2 sig figs. Hope that helps! :)
