Hello :)
Based on the information I received reading the picture, the answer should be “B”
Explanation: if I am wrong I’m very sorry. But that should be the answer
Answer:
I think it will option B it will retain enough heat
Answer:
So the molar mass of C4,H10 is
58.12g mole -1
Answer:
solubility of X in water at 17.0 
is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0 can be calculated using the information given.
Let's assume solubility of X in water at 17.0 is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y = 
= 0.11
Hence solubility of X in water at 17.0 is 0.11 g/mL.
Answer:
[H⁺] = 6.083x10⁻⁴ M, [C₆H₅OO⁻] = 6.083x10⁻⁴ M, [C₆H₅OOH] = 3.98x10⁻³M, pH = 3.22
Explanation:
Data: we have 0.56 gr of benzoic acid, disolved in 1Lt of water. Kₐ = 6.4x10⁻⁵
M (molar mass) of BA (Benzoic Acid) = 122 g/mol
Then, the inicial concentration is 0.56/122 = 4.59x10⁻³ M
We should consider the equation once it reaches the equilibrium:
C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺
C - x x x
And, for the Kₐ:
Kₐ = [H⁺][C₆H₅COO⁻]/[C₆H₅COOH] = x²/(C-x) , where C = 4.59x10⁻³
Then: x² + Kₐx - KₐC = 0
x² + 6.4x10⁻⁵ - 2.9x10⁻⁷ = 0
Resolving this cuadratic equation (remember to use Baskara equation), we obtain:
x = 6.083x10⁻⁴ M
Then: [H⁺] = [C₆H₅COO⁻] = 6.083x10⁻⁴ M
[C₆H₅COOH] = C - x = 3.98x10⁻³ M
pH = -Log [H⁺] = 3.22
