When a substance is heated, it gains thermal energy. Therefore, its particles move faster and its temperature rises. When a substance is cooled, it loses thermal energy, which causes its particles to move more slowly and its temperature to drop.
Answer:
Cupid nitrate is what I'm going for
A is Ea, which stands for activating energy. Energy is needed to get the reaction underway and Ea is the energy needed to “start” the reaction.
B is the temperature either released or absorbed.
The diagram shows that the reaction is exothermic based on the fact that the products energy is lower than the reactants. That is because energy (which is temperature in this case) is released during the process. If the reactants would have been lower than the products, the reaction would be endothermic.
Answer:
74,67 gr/mol
Explanation:
At STP 1 mole of an ideal gas has volume of 22,4 L. Since we know the volume of the gas we can find the number of moles of the gas. (300 mL=0,3 L)
n=0,3L/22,4 L=0,01339 mol
Since we know weight of the gas as 1 g, we can find the molecular weight as;
MW=1 g/0,01339 mol =74,67 gr/mol
Answer:
3,964 years.
Explanation:
- It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
- Half-life time is the time needed for the reactants to be in its half concentration.
- If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
- Also, it is clear that in first order decay the half-life time is independent of the initial concentration.
- The half-life of the element is 5,730 years.
- For, first order reactions:
<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>
Where, k is the rate constant of the reaction.
t1/2 is the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.
- Also, we have the integral law of first order reaction:
<em>kt = ln([A₀]/[A]),</em>
where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of the sample ([A₀] = 100%).
[A] is the remaining concentration of the sample ([A] = 61.9%).
∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.
