1. Convert gallons to mL. 1 gal = 3785.4117840007 mL, multiply that by 29 and get 109776.94173602 mL.
2. Since there is one gram per every mL, there are 109776.94173602 g of water in the fish tank.
3. Convert g to pounds. 1 g = 0.0022 pounds. Multiply 109776.94173602 by 0.0022 and end up with about 241.5 pounds of water.
Answer:
ΔS = -661.0J/mol is the entropy change for the system
ΔS = -842J/mol.K is the entropy change for the surroundings
Explanation:
From the relationship between ΔG, T, ΔH and ΔS,
Mathematically, ΔG = ΔH - TΔS
TΔS = ΔH - ΔS
ΔS = ΔH - ΔS / T
but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)
plugging into the equation,
ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298
ΔS = -0.6610KJ/mol or in J.mol
ΔS = -661.0J/mol is the entropy change for the system
- For entropy change for the surroundings = ΔS = ΔH/T
- ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings
The answer is c because of going left to right
Answer: spring tides
When earth, moon, and sun line up cause high and low tides :)
Answer:
5.0 x 10⁹ 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 K-40 = 1.251 × 10⁹ 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/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ 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 = 5.54 x 10⁻¹⁰ year⁻¹).
t is the time of the reaction (t = ??? year).
[A₀] is the initial concentration of (K-40) ([A₀] = 100%).
[A] is the remaining concentration of (K-40) ([A] = 6.25%).
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))
∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.
∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.
