Chemistry, 20.09.2020 17:01 nev322

A common issue in organic chemistry processes is trace contamination. Organic solvents can become contaminated with water. One method of removing the water contamination is to freeze dry the solvent. Unfortunately, not all solvents, like acetonitrile, are compatible with this method. Another method is to add a chemical to react with the water. One such chemical is calcium hydride, which will react to form calcium hydroxide and hydrogen gas. When you checked the chemical inventory, you realized you only had 24.6g of calcium hydride on hand. If you had a 1.5L container of organic solvent that contained 14.0g of water, what will be the outcome of your attempt to dewater the solvent?

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Answer from: leslie4759

Complete Question

A common issue in organic chemistry processes is trace contamination. Organic solvents can become contaminated with water. One method of removing the water contamination is to freeze dry the solvent. Unfortunately, not all solvents, like acetonitrile, are compatible with this method. Another method is to add a chemical to react with the water. One such chemical is calcium hydride, which will react to form calcium hydroxide and hydrogen gas. When you checked the chemical inventory, you realized you only had 24.6g of calcium hydride on hand. If you had a 1.5L container of organic solvent that contained 14.0g of water, what will be the outcome of your attempt to dewater the solvent?

A ) There is no way to tell

B) CaH2 will be left over

C water will be left over

D Both will be left over

The correct option is B

Explanation:

From the question we are told that

The mass of calcium hydride present is $m_h = 24.6 \ g$

The mass of water is $m_w = 14.0 \ g$

Generally the molar mass of calcium hydride is $Z_h = 42 \ g/mol$

The molar mass of water is $Z_w = 18 \ g / mol$

The reaction between water and calcium hydride can be represented as a balanced chemical equation as follows

$CaH_2 + 2 H_2O \to Ca(OH)_2 + 2H_2$

From this equation we see that

1 mole(42 g) of calcium hydride dries 2(2 * 18 g) moles of water

Then we can say that

if 42 grams of calcium hydride dries (2 * 18 )g of water

24.6 grams will dry z grams of water

=> $z = \frac{24.6 * 2 *18}{42}$

=> $z = 21.1 \ g$ of water

Given that the amount of water the 24.6g of calcium hydride is capable of drying is greater than the amount of water present then the it means that after drying the water available that the calcium hydride will remain

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

$^{m}_{4}x$ and $^{m}_4y$

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

$^{m}_{5}x^{-}$

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be $^{9}_{5}be^{-}$ and $^{9}_{4}be$ .

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.

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