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
Answer: A = 1560
B = 1.6
Explanation: brainlest please
Answer:
The final temperature is 39.58 degree Celsius
Explanation:
As we know
Q = m * c * change in temperature
Specific heat of water (c) = 4.2 joules per gram per Celsius degree
Substituting the given values we get -
5750 = 335 * 4.2 * (X - 35.5)
X = 39.58 degree Celsius
You didn’t show the cylinder containing water, so I created one that you can use as a model (see image).
The water level was originally at 37 mL.
Then you added the ball, and it displaced its volume of water.
The new volume reading is 52 mL, so
Volume of ball = volume of displaced water = 52 mL – 37 mL = 15 mL.