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3 years ago

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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that it is cooler than the lithosphere.

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4 years ago

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Explanation:

5.00 mol H2O × (1 mol C6H12O6/6 mol H2O)

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3 years ago

A chunk of sulfur has a volume of 8.33 cm3. What is the mass of this sulfur? (Density of sulfur = 2.07 g/cm3.)

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Explanation:

equations to note:

density= mass/volume

mass= volume *density

volume= mass/density

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you have a density- 2.07 g/cm3

Answer:

8.33cm3 * 2.07g/cm3= 17.24g

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3 years ago

A wooden tool is found at an archaeological site. Estimate the age of the tool using the following information: A 100 gram sampl

lord [1]

Answer:

2578.99 years

Explanation:

Given that:

100 g of the wood is emitting 1120 β-particles per minute

Also,

1 g of the wood is emitting 11.20 β-particles per minute

Given, Decay rate = 15.3 % per minute per gram

So,

Concentration left can be calculated as:-

C left = $[A_t]=\frac{11.20\ per\ minute}{15.3\ per\ minute\ per\ gram}\times [A_0]= 0.7320[A_0]$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Also, Half life of carbon-14 = 5730 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{5730}\ years^{-1}$

The rate constant, k = 0.000120968 year⁻¹

Time =?

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

So,

$\frac {[A_t]}{[A_0]}=e^{-0.000120968\times t}$

$\frac {0.7320[A_0]}{[A_0]}=e^{-0.000120968\times t}$

$0.7320=e^{-0.000120968\times t}$

$ln\ 0.7320=-0.000120968\times t$

t = 2578.99 years

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3 years ago

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Answer:

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3 years ago