Answer : The value of 
for 
is 
.
Solution : Given,
Solubility of in water = 
The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.
The equilibrium equation is,
Initially - 0 0
At equilibrium - s s
The Solubility product will be equal to,
![K_{sp}=[Ba^{2+}][CO^{2-}_3]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BBa%5E%7B2%2B%7D%5D%5BCO%5E%7B2-%7D_3%5D)
![[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D%5BCO%5E%7B2-%7D_3%5D%3Ds%3D4.4%5Ctimes%2010%5E%7B-5%7Dmole%2FL)
Now put all the given values in this expression, we get the value of solubility constant.
Therefore, the value of for is .
Answer:
pH of HNO₃ having an hydrogen ion concentration of 0.71M is 0.149
Explanation:
HNO₃ (aqueous) ⇄ H⁺ + NO3⁻
The pH is defined as the negative log of the hydrogen ion concentration
pH = - log [H⁺]
From the question, the hydrogen ion concentration is given as 0.71M, therefore
pH = -log [0.71]
= 0.149
Answer: If you think about it, B. would be the most reasonable answer with the given factors.
Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
