Answer: orderly arrangement of particles
Explanation:
In a crystaline solid, the particles that compose the solid are arranged or packed in an orderly manner to form a three dimensional crystal lattice with a defined structure. Sodium chloride is a crystaline solid.
In an amorphous solid, the particles that compose the solid aren't arranged in an orderly manner hence the solid tends to be brittle, e.g glass
Answer:
Eºcell = -1.78 V
Explanation:
The Eº cell is an intensive property, i.e they do not depend on the quantity of material present and the desired reaction in our problem is exactly half the reverse of the one given, the Eºcell will then be the negative of the first then Eºcell is -1.78 V and the redox reaction will be non-spontaneous as opposed to the first.
Answer:
1. 0.97 V
2. 
Explanation:
In this case, we can start with the <u>half-reactions</u>:


With this in mind we can <u>add the electrons</u>:
<u>Reduction</u>
<u>Oxidation</u>
The reduction potential values for each half-reaction are:
- 0.69 V
-1.66 V
In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:
+1.66 V
Finally, to calculate the overall potential we have to <u>add</u> the two values:
1.66 V - 0.69 V = <u>0.97 V</u>
For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

I hope it helps!
Not sure about the last one, but I think the percent composition of zinc would be ~ 42.8232
pH of the buffer solution is 1.76.
Chemical dissociation of formic acid in the water:
HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)
The solution of formic acid and formate ions is a buffer.
[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions
[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate
[HCOOH] = 1.45 M - 0.015 M
[HCOOH] = 1.435 M; equilibrium concentration of formic acid
pKa = -logKa
pKa = -log 1.8×10⁻⁴ M
pKa = 3.74
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)
pH = 3.74 + log (0.015 M/1.435 M)
pH = 3.74 - 1.98
pH = 1.76
More about buffer: brainly.com/question/4177791
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