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3 years ago

Identify the number of moles in 369 grams of calcium hydroxide. Use the periodic table and the polyatomic ion resource.

Chemistry

1 answer:

topjm [15]3 years ago

7 0

Answer : The number of moles in 369 grams of calcium hydroxide is, 4.98 moles

Explanation : Given,

Mass of calcium hydroxide = 369 g

Molar mass of calcium hydroxide = 74.093 g/mole

Formula used :

$\text{Moles of calcium hydroxide}=\frac{\text{Mass of calcium hydroxide}}{\text{Molar mass of calcium hydroxide}}$

Now put all the given values in this formula, we get the moles of calcium hydroxide.

$\text{Moles of calcium hydroxide}=\frac{369g}{74.093g/mole}=4.98mole$

Therefore, the number of moles in 369 grams of calcium hydroxide is, 4.98 moles

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3 years ago

s the purpose of the CaCl2 drying tube? What chemical reaction is it preventing (please supply a mechanism).

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Answer:

See explanation

Explanation:

A drying tube prevents moisture from contaminating the reactants. The drying tube contains CaCl2 a hygroscopic material whose function is to absorb the moisture so that it does not react with the Grignard reagent. Without the desiccating action of CaCl2, moisture will enter the reaction chamber,contaminating the reactants.

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The reaction of equal molar amounts of benzene, C6H6, and chlorine, Cl2, carried out under special conditions, yields a gas and

WINSTONCH [101]

Answer : The molecular formula of a compound is, $C_6H_5Cl_1$

Solution : Given,

Mass of C = 64.03 g

Mass of H = 4.48 g

Mass of Cl = 31.49 g

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Molar mass of H = 1 g/mole

Molar mass of Cl = 35.5 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{64.03g}{12g/mole}=5.34moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4.48g}{1g/mole}=4.48moles$

Moles of Cl = $\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{31.49g}{35.5g/mole}=0.887moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.34}{0.887}=6.02\approx 6$

For H = $\frac{4.48}{0.887}=5.05\approx 5$

For Cl = $\frac{0.887}{0.887}=1$

The ratio of C : H : Cl = 6 : 5 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_6H_5Cl_1$

The empirical formula weight = 6(12) + 5(1) + 1(35.5) = 112.5 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{112.5}{112.5}=1$

Molecular formula = $(C_6H_5Cl_1)_n=(C_6H_5Cl_1)_1=C_6H_5Cl_1$

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3 years ago

Suppose that 0.50 grams of barium-131 are administered orally to a patient. Approximately how many milligrams of the barium woul

nalin [4]

Two months later 13.8 milligrams of the barium-131 still be radioactive.

<h3>How is the decay rate of a radioactive substance expressed ? </h3>

It is expressed as:

$A = A_{0} \times (\frac{1}{2})^{t/T}$

where,

A = Amount remaining

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Here

A₀ = 0.50g

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Now put the values in above expression we get

$A = A_{0} \times (\frac{1}{2})^{t/T}$

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Thus from the above conclusion we can say that Two months later 13.8 milligrams of the barium-131 still be radioactive.

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Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Suppose that 0.50 grams of ban that 0.50 grams of barium-131 are administered orally to a patient. Approximately many milligrams of the barium would still be radioactive two months later? The half-life of barium-131 is 11.6 days.

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djverab [1.8K]

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3 years ago