<h2>Natural Abundance for 10B is 19.60%</h2>
Explanation:
- The natural isotopic abundance of 10B is 19.60%.
- The natural isotopic abundance of 11B is 80.40%.
- The isotopic masses of boron are 10.0129 u and 11.009 u respectively.
For calculation of abundance of both the isotopes -
Supposing it was 50/50, the average mass would be 10.5, so to increase the mass we need a more percentage of 11.
Determining it as an equation -
10x + 11y= 10.8
x+y=1 (ratio)
10x + 10y = 10
By taking the denominator away from the numerator
we get;
y = 0.8
x + y = 1
∴ x = 0.2
To get percentages we need to multiply it by 100
So, the calculated abundance is 80% for 11 B and 20% 10 B.
Answer:
milk, lemon juice, toothpaste and baking soda
Answer:
Chemical indicator, any substance that gives a visible sign, usually by a colour change, of the presence or absence of a threshold concentration of a chemical species, such as an acid or an alkali in a solution. An example is the substance called methyl yellow, which imparts a yellow colour to an alkaline solution.
Explanation:
Answer:
0.210 M
Explanation:
<em>A 75.0 mL aliquot of a 1.70 M solution is diluted to a total volume of 278 mL.</em>
In order to find out the resulting concentration (C₂) we will use the dilution rule.
C₁ × V₁ = C₂ × V₂
1.70 M × 75.0 mL = C₂ × 278 mL
C₂ = 0.459 M
<em>A 139 mL portion of that solution is diluted by adding 165 mL of water. What is the final concentration? Assume the volumes are additive.</em>
Since the volumes are additive, the final volume V₂ is 139 mL + 165 mL = 304 mL. Next, we can use the dilution rule.
C₁ × V₁ = C₂ × V₂
0.459 M × 139 mL = C₂ × 304 mL
C₂ = 0.210 M
