Answer:
The percent isotopic abundance of C- 12 is 98.93 %
The percent isotopic abundance of C- 13 is 1.07 %
Explanation:
we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)
First of all we will set the fraction for both isotopes
X for the isotopes having mass 13.003355
1-x for isotopes having mass 12
The average atomic mass of carbon is 12.0107
we will use the following equation,
13.003355x + 12 (1-x) = 12.0107
13.003355x + 12 - 12x = 12.0107
13.003355x- 12x = 12.0107 -12
1.003355x = 0.0107
x= 0.0107/1.003355
x= 0.0107
0.0107 × 100 = 1.07 %
1.07 % is abundance of C-13 because we solve the fraction x.
now we will calculate the abundance of C-12.
(1-x)
1-0.0107 =0.9893
0.9893 × 100= 98.93 %
98.93 % for C-12.
I think it's Chlorine but, not 100% sure. so C.
Answer:
The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²
Explanation:
The given parameters is as follows;
The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g
The required information = The number of formula units of potassium chloride (KCl)
The Molar Mass of KCl = 74.5513 g/mol
Therefore, we have;
1 mole of a substance, contains Avogadro's number (6.022 × 10²³) of formula units
Therefore;
0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³ ≈ 3.077588 × 10²² formula units
From which we have, the number of formula units in 3.81 g of potassium chloride (KCl) ≈ 3.08 × 10²² formula units.
I think 8 or 6 im not sure
The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles
<h3>Balanced equation </h3>
2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
<h3>How to determine the theoretical yield of NaBr</h3>
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
Therefore,
2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr
Therefore,
Thus, the theoretical yield of NaBr is 7.08 moles
Learn more about stoichiometry:
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