Answer:
Approximately 
, assuming that this acid is monoprotic.
Explanation:
Assume that this acid is monoprotic. Let 
denote this acid.
.
Initial concentration of without any dissociation:
![[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D%20%3D%200.730%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D)
.
After 
of that was dissociated, the concentration of both 
and 
(conjugate base of this acid) would become:
.
Concentration of in the solution after dissociation:
.
Let ![[{\rm HA}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D)
, ![[{\rm H}^{+}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D)
, and ![[{\rm A}^{-}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D)
denote the concentration (in 
or 
) of the corresponding species at equilibrium. Calculate the acid dissociation constant 
for , under the assumption that this acid is monoprotic:
![\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DK_%7B%5Crm%20a%7D%20%26%3D%20%5Cfrac%7B%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D%7D%7B%5B%7B%5Crm%20HA%7D%5D%7D%20%5C%5C%20%26%3D%20%5Cfrac%7B%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%20%5Ctimes%20%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%7D%7B0.63875%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%7D%5C%5C%5B0.5em%5D%26%5Capprox%201.30%20%5Ctimes%2010%5E%7B-2%7D%20%5Cend%7Baligned%7D)
.
Answer:
603000 J
Explanation:
The following data were obtained from the question:
Energy required (Q) =...?
Mass (M) = 10000 g
Specific heat capacity (C) = 2.01 J/g°C
Overheating temperature (T2) = 121°C
Working temperature (T1) = 91°C
Change in temperature (ΔT) =.?
Change in temperature (ΔT) =T2 – T1
Change in temperature (ΔT) = 121 – 91
Change in temperature (ΔT) = 30°C
Finally, we shall determine the energe required to overheat the car as follow:
Q = MCΔT
Q = 10000 × 2.01 × 30
Q = 603000 J
Therefore, 603000 J of energy is required to overheat the car.
<u>Explanation:</u>
It is given that polar solutes can be dissolved in polar solvents and non-polar solutes can be dissolved in non-polar solvent.
Alcohol being polar, does not dissolves ionic salt in it.
is a non-polar solvent.
From the given options:
1. NaCl: This is an ionic salt and hence, it will be soluble in water only.
2. 
: Iodine gas is a non-polar solute and hence, will be dissolved in non-polar solvent which is
3. Ethanol: As, it is a polar molecule and is not an ionic salt, therefore it can be soluble in both water and alcohol.
4. Benzene: It is a non-polar molecule and hence, it will be dissolved in
5. 
: Bromine gas is a non-polar solute and hence, will be dissolved in
6. 
: This is an ionic salt and hence, it will be soluble in water only.
7. Toluene: This is a non-polar solute and hence, will be dissolved in
8. 
: This is an ionic salt and hence, it will be soluble in water only.
Answer:
9.62 × 10² J
Explanation:
In this problem, we need to convert 230 calories to joules. For this, we must know the conversion from calories to joules.
1 Calorie = 4.184 Joules
To find Joules in 230 calories, multiply 4.184 by 23. So,
230 Calorie = (230 × 4.184) Joules
= 962.32 Joules
or
230 Calories = 9.62 × 10² J
So, the correct option is (A).
