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:
It get thicker beacause the mid ocean
Explanation:
Because when it gets moved back the heat rises and it builds up to be thicker.
As we know that
<span>V1/T1 = V2/T2
V1 = 9.10 L
T1 = 471 K
V2 = 2.50 L
T2 = 2.5 x 471 / 9.10 = 129.3 K
T2 = 129.3 - 273 =
-143.6 deg Celsiu
hope it helps</span>
EVERY YEAR, NIKON <span>selects the most artful, scientifically enlightening and skillfully produced images from thousands of submissions for its </span>Small World<span> microscope photography contest. Tomorrow, another set of impressive winners will be announced for the contest’s 40th year.</span>
