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.
Answer:
bombarding it with an energetic particle
Explanation: nuclear reaction, a change in the identity or characteristics of an atomic nucleus, induced by bombarding it with an energetic particle. The bombarding particle may be an alpha particle, a gamma-ray photon, a neutron, a proton, or a heavy-ion.
Answer: The space occupied by the gas at 400 torr and 
is 250 mL.
Explanation:
Given: 
= 250 mL, 
= 800 torr, 
, 
= 400 torr, 
Formula used is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that space occupied by the gas at 400 torr and is 250 mL.
Answer: Fluorine and chlorine are gases. Bromine is one of only two liquid elements, and iodine is a solid.
Explanation:
The given data is as follows.
Concentration = 0.1 
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
= 
T = 
= (30 + 273) K = 303 K
Formula for electric double layer thickness (
) is as follows.
= 
where, 
= concentration =
Hence, putting the given values into the above equation as follows.
= 
= 
= 
m
or, = 
= 1 nm (approx)
Also, it is known that = 
Hence, we can conclude that addition of 0.1 of KCl in 0.1 of NaBr "" will decrease but not significantly.
