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vovikov84 [41]
3 years ago
14

Q3. A student added water to calcium oxide to make calcium hydroxide

Chemistry
1 answer:
galina1969 [7]3 years ago
6 0

26.4 kg of calcium hydroxide (Ca(OH)₂)

Explanation:

We have the following chemical reaction:

CaO + H₂O → Ca(OH)₂

molecular weight of Ca(OH)₂ = atomic mass of Ca × number of Ca atoms + atomic mass of O × number of O atoms + atomic mass of H × number of H atoms

molecular weight of Ca(OH)₂ = 40 × 1 + 16 × 2 + 1 × 2 = 74 g/mole

molecular weight of CaO = atomic mass of Ca × number of Ca atoms + atomic mass of O × number of O atoms

molecular weight of CaO = 40 + 16 = 56 g/mole

number of moles = mass / molecular weight

number of moles of CaO = 20 / 56 = 0.357 kmoles

Taking in account the chemical reaction we devise the following reasoning:

if        1 kmole of CaO produces 1 kmole of Ca(OH)₂

then   0.357  kmoles of CaO produces X kmoles of Ca(OH)₂

X = (0.357 × 1) / 1 = 0.357 kmoles of Ca(OH)₂

mass = number of moles × molecular weight

mass of Ca(OH)₂ = 0.357 × 74 = 26.4 kg

Learn more about:

number of moles

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If two atoms are isotopes of the same element the atoms must have
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Same number of protons but different number of neutrons

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The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be
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Explanation:

1) Initial mass of the Cesium-137=N_o= 180 mg

Mass of Cesium after time t = N

Formula used :

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

Half life of the cesium-137 = t_{1/2}=30 years[p/tex]where,&#10;[tex]N_o = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

t_{\frac{1}{2}} = half life of the isotope

\lambda = rate constant

N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}

Now put all the given values in this formula, we get

N=180mg\times e^{-\frac{0.693}{30 years}\times t}

Mass that remains after t years.

N=180 mg\times e^{0.0231 year^{-1}\times t}

Therefore, the parent isotope remain after one half life will be, 100 grams.

2)

t = 70 years

N_o=180 mg

t_{1/2}= 30 yeras

N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

3)

N_o=180 mg

t_{1/2}= 30 yeras

N = 1 mg

t = ?

1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}

\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0
3 years ago
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