<span>(A)hydrochloric acid + silver nitrate
HCl(aq) + AgNO3(aq) -----> AgCl(s) +HNO3(aq)
</span><span>(B)hydrochloric acid + sodium hydroxide
</span><span>HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)
</span><span>(C)calcium chloride + silver nitrate
CaCl2(aq) + AgNO3(aq) ----> </span>AgCl(s) +Ca(NO3)2(aq)
<span>(D)sodium chloride + silver nitrate
</span>NaCl(aq) + AgNO3(aq) ----> AgCl(s) +NaNO32(aq)
AgCl is a white precipitate.
In (B) no precipitate was formed, so answer is B.
Answer : The expected coordination number of NaBr is, 6.
Explanation :
Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.
This is represented by,
When the radius ratio is greater than 0.155, then the compound will be stable.
Now we have to determine the radius ration for NaBr.
Given:
Radius of cation, 
= 102 pm
Radius of cation, 
= 196 pm
As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.
The relation between radius ratio and coordination number are shown below.
Therefore, the expected coordination number of NaBr is, 6.
<span>If the aqueous solution is 34% Licl then it is 100 - 34% water = 66%
From the calculation we've found out that it is 66% water. Then we need to find the weight from a 250 g solution.
66/100 * 250 = 165g
Hence it is 165g</span>
Answer:
the overall cell potential
Explanation:
We must bear in mind that the standard hydrogen electrode is a reference electrode whose electrode potential has been arbitrarily set at 0 V.
The standard hydrogen electrode consists of hydrogen ion solution and hydrogen gas together with a platinum electrode.
The overall cell potential is the reduction potential of the substance being determined using the standard hydrogen electrode as a reference electrode since its electrode potential is set at zero volts.
