Answer:
41 g
Explanation:
We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.
pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]
pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]
log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]
log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40
[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M
We can find the mass of NaC₆H₅COO using the following expression.
M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L
mass NaC₆H₅COO = 41 g
solution:
1000 = m*2400*(78-22) + m*8.79*10^5
1000= 134400m + 879000m
1000= 1030200m
m = 1000/1013400
m= 1013.4 grams
the final answer is 0.9706 grams
Answer:
A. 32.6 g/mol
Explanation:
First convert the volume of gas to moles using the ratio 1 mol / 22.4 L at STP.
0.070 L • (1 mol / 22.4 L) = 0.00313 mol
Now divide the grams of gas by the moles of gas:
0.102 g / 0.00313 mol = 32.6 g/mol
Answer:
To determine the amount of heat the gold has absorbed to melt, we simply multiply the mass of the block of ice to the heat of fusion of water which is given above. We calculate as follows:
Heat = 20.0 g (35.4 g)
Heat = 1290 J
Nor liquid or solid since flubber is like oobleck, it is a non newtonian object.