The activity of the sample when it was shipped from the manufacturer is 4.54 mCi
<h3>How to determine the number of half-lives that has elapsed </h3>
From the question given above, the following data were obtained:
- Time (t) = 48 hours
- Half-life (t½) = 14.28 days = 14.28 × 24 = 342.72 hours
- Number of half-lives (n) =?
n = t / t½
n = 48 / 342.72
n = 0.14
<h3>How to determine the activity of the sample during shipping </h3>
- Number of half-lives (n) = 0.14
- Original activity (N₀) = 5.0 mCi
- Activity remaining (N) =?
N = N₀ / 2ⁿ
N = 5 / 2^0.14
N = 4.54 mCi
Thus, the activity of the sample during shipping is 4.54 mCi
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Gain .....
It loses electron when a positive charge is formed..ex Na+
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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The most appropriate answer is B !!
Covalent bond is formed by sharing of electrons !!
Answer:
Actual yield = 86.5g
Explanation:
Percent yield = 82.38%
Theoretical yield = 105g
Actual yield = x
Equation of reaction,
CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O
Percentage yield = (actual yield / theoretical yield) * 100
82.38% = actual yield / theoretical yield
82.38 / 100 = x / 105
Cross multiply and make x the subject of formula
X = (105 * 82.38) / 100
X = 86.499g
X = 86.5g
Actual yield of CaCl₂ is 86.5g
