Which describes the titration graph when a strong acid is titrated by a weak base?
1 answer:
You might be interested in
Answer:
Explanation:
We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.
We will use the following formula to calculate heat energy.
The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.
- ΔT = final temperature - inital temperature
The aluminum block was heated from 23.0 °C to 73.5 °C.
- ΔT= 73.5 °C - 23.0 °C = 50.5 °C
Now we know all three variables and can substitute them into the formula.
- m= 225 g
- c= 0.897 J/g° C
- ΔT= 50.5 °C
Multiply the first two numbers. The units of grams cancel.
Multiply again. This time, the units of degrees Celsius cancel.
The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.
Multiply by the answer we found in Joules.
The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.
Approximately <u>10.2 kilojoules</u> of energy would be required.
Answer:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
.
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both and
will increase if the pressure is increased through compression. However, because
and
have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient
.
As a result, the increase in pressure will have no impact on the value of . If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.