Answer:
Im guessing it's C....................................................
Answer:
Half life is 6 years.
Explanation:
T½ = In2 / λ
Where λ = decay constant.
But N = No * e^-λt
Where N = final mass after a certain period of time
No = initial mass
T = time
N = 0.625g
No = 10g
t = 24 years
N = No* e^-λt
N / No = e^-λt
λ = -( 1 / t) In N / No (inverse of e is In. Check logarithmic rules)
λ = -(1 / 24) * In (0.625/10)
λ = -0.04167 * In(0.0625)
λ = -0.04167 * (-2.77)
λ = 0.1154
T½ = In2 / λ
T½ = 0.693 / 0.1154
T½ = 6.00 years.
The half life of radioactive cobalt-60 is 6 years
Answer:
Explanation:
The pressure of a gaseous mixture is equal to the sum of the partial pressures of the individual gases:
Σ
The prompt is trying to confuse you, but it actually tells us the pressure of the mixture to be 1 atm, but this can be converted to torr. Furthermore, we are informed only three gases are in the mixture: diatomic nitrogen, diatomic oxygen, and carbon dioxide:
Solve for Po2:
Thus, the partial pressure of diatomic oxygen is 177.707 torr.
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Answer:
2Cu2S + 3O2 + 2C -------> 4Cu + 2SO2 + 2CO
Explanation:
Equation 1 should correctly be written as;
2Cu2S + 3O2-----> 2Cu2O + 2SO2
Equation 2 should be correctly written as;
2Cu2O + 2C -----> 4Cu + 2CO
The overall reaction equation is;
2Cu2S + 3O2 + 2C -------> 4Cu + 2SO2 + 2CO
Note that species that are intermediates are cancelled out .
Answer:
No, ΔE does not always equal zero because it refers to the systems internal energy, which is affected by heat and work
Explanation:
According to the first law of thermodynamics, energy is neither created nor destroyed. This implies that the total energy of a system is always a constant.
So, according to the first law of thermodynamics we have that ΔE = q + w. This means that the value of ΔE depends on q (heat) and w(work). Hence ΔE is not always zero since it depends on the respective values of q and w.
