Answer:
18.94%.
Explanation:
- The decay of carbon-14 is a first order reaction.
- The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.
- The integration law of a first order reaction is:
<em>kt = ln [A₀]/[A]</em>
k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.
t is the time = 13,750 years.
[A₀] is the initial percentage of carbon-14 = 100.0 %.
[A] is the remaining percentage of carbon-14 = ??? %.
∵ kt = ln [Ao]/[A]
∴ (1.21 x 10⁻⁴ year⁻¹)(13,750 years) = ln (100.0%)/[A]
1.664 = ln (100.0%)/[A]
Taking exponential for both sides:
5.279 = (100.0%)/[A]
<em>∴ [A]</em> = (100.0%)/5.279 = <em>18.94%.</em>
Answer: (22.98977 g Na/mol) + (1.007947 g H/mol) + (12.01078 g C/mol) + ((15.99943 g O/mol) x 3) = 84.0067 g NaHCO3/mol
9.
(1.20 g NaHCO3) / (84.0067 g NaHCO3/mol) = 0.0143 mol NaHCO3
10.
Supposing the question is asking about "how many moles" of CO2. And supposing the reaction to be something like:
NaHCO3 + H{+} = Na{+} + H2O + CO2
(0.0143 mol NaHCO3) x (1 mol CO2 / 1 mol NaHCO3) = 0.0143 mol CO2 in theory
11.
n = PV / RT = (1 atm) x (0.250 L) / ((0.0821 L atm/K mol) x (298 K)) = 0.0102 mol CO2
12.
(0.0143 mol - 0.0102 mol) / (0.0143 mol) = 0.287 = 28.7%
Explanation:
Answer:
.0556 L
Explanation:
First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.
Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).
Finally, the answer will be .0556 L.
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Nuclear fission is the process of which a large nucleus splits into two smaller nuclei with the release of energy and neutron. In simpler words, nuclear fission is the process in which a nucleus is split into two smaller fragments or pieces (nuclei) and so energy and neutrons are released. The resulting pieces of this fission process have less combined mass than the original piece (nucleus) and the missing was is converted into nuclear energy.
