Answer : The final concentration of
is, 2.9 M
Explanation :
Expression for rate law for first order kinetics is given by:

where,
k = rate constant = 
t = time passed by the sample = 3.5 min
a = initial concentration of the reactant = 3.0 M
a - x = concentration left after decay process = ?
Now put all the given values in above equation, we get


Thus, the final concentration of
is, 2.9 M
Answer:
b- The heat capacity ratio increases but output temperature don’t change
Explanation:
The heat capacity is the amount of energy required to raise the temperature of a body, by 1 degree. On the other hand, the specific heat capacity is the amount of heat required to raise the temperature of a of unit mass of a material by 1 degree.
Heat capacity is an extensive property meaning its value depends on the amount of material. Specific heat capacity is found by dividing heat capacity by the mass of the sample, thus making it independent of the amount (intensive property). So if the specific heat capacity increases and the mass of the sample remains the same, the heat capacity must increase too. Because of that options c and d that say that heat capacity reamins same are INCORRECT.
On the other hand, in which has to be with options a and b both say that the heat capacity increases which is correct, but about the output temperatures what happens is that if we increase the specific heat capacity of both fluids that are involved in a process of heat exchange in the same value, the value of the output temperatures do not change so only option a is CORRECT.
Well you should study how different chemicals work, and make flashcards to carry around or make notes during class to study for later
Answer:

Explanation:
Step 1. Determine the cell potential
<u> E°/V </u>
2×[Cr ⟶ Cr³⁺ + 3e⁻] 0.744 V
<u>3×[Cu²⁺ + 2e⁻ ⟶ Cu] </u> <u>0.3419 V
</u>
2Cr + 3Cu²⁺ ⟶ 3Cu + 2Cr³⁺ 1.086 V
Step 2. Calculate ΔG°

is the Calcium
because the calcium in group two which group two has 2 electron and also calcium and Potassium 4s sub level group potassium has 1 electron. :)