Using the exponential decay model; we calculate "k"
We know that "A" is half of A0
A = A0 e^(k× 5050)
A/A0 = e^(5050k)
0.5 = e^(5055k)
In (0.5) = 5055k
-0.69315 = 5055k
k = -0.0001371
To calculate how long it will take to decay to 86% of the original mass
0.86 = e^(-0.0001371t)
In (0.86) = -0.0001371t
-0.150823 = -0.0001371 t
t = 1100 hours
The mass decay rate is of the form
where
m₀ = 3000 g,the initial mass
k = the decay constant
t = time, years.
Because the half-life is 30 years, therefore
After 60 years, the mass remaining is
Answer: 750 g
Answer: The Answer is C.
Explanation: The Nucleus only makes up less than .01% of the volume of the Atom. The Nucleus does contain more than 99.9% of the mass of the Atom. I hope that this helps you! Good luck!
Answer:
through natural selection
Explanation:
Natural selection is the driving mechanism of the evolution of species. Organisms can survive more in an environment if they are capable to adapt the changing environmental conditions. After surviving they are also able to pass their modified trait into their future generations (evolution).
Helpful variations accumulate through natural selection, as the organisms that are better adapted to their environment reproduce offspring with the same variations.
The volume of the NaOH used is calculated as 14 mL.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with the calculation of the amount of substance in a reaction using mass - mole or mass - volume relationship.
Here;
Number of moles of CaCO3 = 0.205 g/100.1 = 0.00205 moles
Number of moles of HCl = 2.00 M * 7/1000 L = 0.014 moles
2 moles of HCl reacts with 1 mole of CaCO3
x moles of HCl reacts with 0.00205 moles of CaCO3
x = 0.00205 moles * 2/1 = 0.0041 moles
Hence HCl is the excess reactant
Amount of excess HCl = 0.014 moles - 0.0041 moles = 0.0099 moles
Concentration of excess HCl reacted = 0.0099 moles/125 * 10^-3 = 0.0792 M
Using;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
VB = CAVANB/CBNA
VB = 0.0792 M * 10 mL * 1/ 0.058 M
VB = 14 mL
Missing parts;
A 0.205 g sample of caco3 (mr = 100.1 g/mol) is added to a flask along with 7.50 ml of 2.00 m hcl. caco3(aq) + 2hcl(aq) → cacl2(aq) + h2o(l) + co2(g) enough water is then added to make a 125.0 ml solution. a 10.00 ml aliquot of this solution is taken and titrated with 0.058 m naoh. naoh(aq) + hcl(aq) → h2o(l) + nacl(aq) how many ml of naoh are used?
Learn more about stoichiometry: brainly.com/question/9743981
