The atomic number is always equal to the number of protons. It will be equal to the number of elections also if the atom has a neutral charge.
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A solid that forms and separates from a liquid mixture is a chemical change.
1.
V = 200 mL (volume)
c = 3 M = 3 mol/L (concentration)
First we convert mL to L:
200 mL = 0.2 L
Then we calculate the moles using the formula: n = V × c = 0.2 L × 3 mol = 0.6 mol
Finally, we just use the molar mass of CaF2 to calculate the actual mass:
molar mass = 78 g/mol
The formula is: m = n × mm (mass = moles × molar mass)
m = 0.6 mol × 78 g/mol = 46.8 g
2.
For this question the steps are exactly like the first question.
V = 50mL = 0.05 L
c = 12 M = 12 mol/L
n = V × c = 0.05 L × 12 mol/L = 0.6 mol
molar mass (HCl) = 36.5 g/mol
m = n × mm = 0.6 mol × 36.5 g/mol = 21.9 g.
3.
The steps for this question are the opposite way.
m(K2CO3) = 250 g
molar mass = 138 g/mol
n = m ÷ mm = 1.81 mol
c = 2 mol/L
V = n ÷ c = 1.81 mol ÷ 2 mol/L = 0.905 L = 905 mL
<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:
The splash cause the volume of the block to be too high
Explanation:
This is because, from the previous measurement made by the student, the volume of the block is 5.7ml higher.
