Answer:
B. Q > K precipitate will form
Explanation:
The reaction is;
Ba(NO3)2(aq) + Na2CO3(aq) ------> BaCO3(s) + 2NaNO3(aq)
Hence the reaction could form a precipitate of BaCO3.
Number of moles of carbonate ions = 50/1000 * 0.10 M = 5 * 10^-3 moles
Number of moles of Barium ions = 20/1000 * 0.10 M = 2 * 10^-3 moles
Total volume after reaction = 20ml + 50ml = 70 ml or 0.07 L
Molarity Barium ions = 5 * 10^-3 moles/ 0.07 L = 0.07 M
Molarity carbonate ions = 2 * 10^-3 moles/ 0.07 L =0.03 M
Q = [Ba^2+] [CO3^2-] = 0.07 * 0.03 = 2.1 * 10^-3
But K = 2.58 × 10
^−
9
We can clearly see that Q>K therefore precipitate will form
The correct answer for the given question above would be option B. HYDROGEN. The form that has the greatest number of <span>strong covalent bonds and is therefore the backbone of organic molecules including carbohydrates, proteins, lipids and nucleic acids is HYDROGEN. Hope this answers your question.</span>
Answer:
The value of entropy change for the process 
Explanation:
Mass of the ideal gas = 0.0027 kilo mol
Initial volume 
= 4 L
Final volume 
= 6 L
Gas constant for this ideal gas ( R ) = 
Where 
= Universal gas constant = 8.314 
⇒ Gas constant R = 8.314 × 0.0027 = 0.0224 
Entropy change at constant temperature is given by,
Put all the values in above formula we get,
![dS = 0.0224 log _{e} [\frac{6}{4}]](https://tex.z-dn.net/?f=dS%20%3D%200.0224%20%20log%20_%7Be%7D%20%5B%5Cfrac%7B6%7D%7B4%7D%5D)
This is the value of entropy change for the process.
