Determine the density of gold if a 475.09g sample occupy a space of 24.6cm^3
1 answer:
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Answer:
Explanation:
The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.
1 .Calculate the hydronium ion concentration
We can use an ICE table to organize the calculations.
HF + H₂O ⇌ H₃O⁺ + F⁻
I/mol·L⁻¹: 2.7 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 2.7 - x x x
2. Calculate the pH
3. Calculate [C₆H₅O⁻]
C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺
2.7 x 0.0441
Answer:
1 - Salts required that can be compared directly, using Ksp values, to estimate solubilities for copper (II) sulfide are - (Option b) ,
(Option c) and
(Option d)
2 - Salts that can be compared directly, using Ksp values, to estimate solubilities for zinc hydroxide are - (Option a)
Explanation:
values can be used to compare the solubility of salts that produce ions in a 1:1 ratio.
{copper (II)sulfide} dissociates with ions in ratio 1:1 .
Because the salts ,
and
have a 1:1 ion ratio, the solubilities of options b, c, and d can be compared to salt 1.
Values can be used to compare the solubility of salts that produce ions in a 1:2 or 2:1 ratio.
2. dissociates into ions in ratio 1:2 .
Because the ions in the salt are in a 1:2 ratio, the solubility of salt 2 can be compared to that of salt 'a'.
Hence , Salt 1 is matched with options b , c , d.
Salt 2 is matched with option a.