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
B. All atoms of silver have the same atomic number but different numbers of neutrons in the nucleus.
There are 12<span> protons and </span>10<span> electrons in a </span><span>Mg<span>2+</span></span><span> ion, the normal amount of neutrons is </span>12<span>.</span>
Using a calculator:
(2.568 x 5.8)/4.186 = 3.5581460…
= 3.56 (3sf)
You didn’t specify the correct number of significant figures needed.
Answer: There are five significant figures in 865,010.
Explanation:
When a degree of accuracy is stated by each digit present in a mathematical figure then it is called a significant figure.
Rules for counting significant figures is as follows.
- Any non-zero digits and zeros present between a non-zero figure are counted. For example, 3580009 has seven significant figures.
- Trailing zeros are counted in a non-zero figure. For example, 0.00250 has three significant figures.
- Leading zeros are not counted. For example, 0.0025 has two significant figures.
So, in the given figure 865010 has five significant figures and the trailing zero will not be counted.
Thus, we can conclude that there are five significant figures in 865,010.
