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2 years ago

Use average bond energies to calculate ΔHrxn for the following hydrogenation reaction: H2C=CH2(g)+H2(g)→H3C−CH3(g)

Chemistry

1 answer:

marissa [1.9K]2 years ago

4 0

Answer:

The $\Delta H_{rxn}$ of the given reaction is -129.6 kJ

Explanation:

The given chemical reaction is as follows.

$H_{2}C=CH_{2}(g)+H_{2}(g)\rightarrow H_{3}C-CH_{3}(g)$

Enthalpy of each reactant and products are as follows.

$\Delta H_{C=C}\,=615.0\,kJ\,mol^{-1}$

$\Delta H _{H-H}\,=435.1\,kJ\,mol^{-1}$

$\Delta H _{C-C}\,=347.3\,kJ\,mol^{-1}$

$\Delta H _{C-H}\,=416.2\,kJ\,mol^{-1}$

In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.

$\Delta H_{rxn}=-\Delta H_{C-C}-2\Delta H_{C-H}+\Delta H_{C=C}+\Delta H_{H-H}$

$=-347.4-2\times416.2+615.0+435.1$

$=-129.6 \,kJ$

Therefore, The $\Delta H_{rxn}$ of the given reaction is -129.6 kJ

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A solution is made by dissolving 9.74 g of sodium sulfate in water to a final volume of 165 mL of solution. What is the weight/w

MatroZZZ [7]

Answer: The weight/weight % or percent by mass of the solute is 5.41 %.

Explanation:

Mass of the sodium sulfate,w = 9.74 g

Volume of the water = 165 mL

Density of the water = 1 g/mL

$Density=1 g/mL=\frac{\text{Mass of water}}{\text{Volume of water}}$

Mass of the water = $1 g/mL\times 165 mL=165 g$

Mass of the solution, W:

Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g

$w/w\%=\frac{w\times 100}{W}=\frac{9.45 g\times 100}{174.47 g}=5.41 \%$

The weight/weight % or percent by mass of the solute is 5.41 %.

8 0

2 years ago

A sample of an unknown compound was decomposed and found to be composed of 1.36 mol oxygen, 4.10 mol hydrogen, and 2.05 mol carb

Flura [38]

Answer:

C3H6O2

Explanation:

To find the empirical formula of the compound, we divide the amount in moles of each of the elements by the amount in mole of the element with the smallest number of mole. In this question, the element with the smallest number of moles is oxygen with 1.36 mole. Hence, we divide the number of moles of each element by this.

H = 4.10/1.36 = 3

O = 1.36/1.36 = 1

C = 2.05/1.36 = 1.5

We then multiply through by 2 to yield the compound with the empirical formula C3H6O2

7 0

2 years ago

A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

elena55 [62]

Answer:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Explanation:

Mass of an alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Mass of water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

From energy balance equation

Heat lost by alloy = Heat gain by water

$m_{a}$ $C_{a}$ [ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This is the specific heat of the alloy.

4 0

2 years ago

What would have happened to your results if during the dehydration some of the copper (ii) sulfate splatter out of the crucible-

lidiya [134]

Answer : The results would show more amount of water in the hydrated sample.

Explanation :

The amount of water of crystallization can be found by taking the masses of hydrated copper sulfate and anhydrous copper sulfate.

The difference in masses indicates the mass of water lost during dehydration process.

If during dehydration process, some of the copper sulfate spatters out of the crucible, then this would give us less mass for anhydrous sample than the actual.

As a result, the difference in masses of hydrated sample and the anhydrous sample would be more.

Therefore the results would show more amount of water in the hydrated sample.

4 0

2 years ago

The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is __________% by

amm1812

Answer:

The concentration of sodium chloride in an aqueous solution that is 2.23 M and that has a density of 1.01 g/mL is 12.90% by mass

Explanation:

2.23 M aqueous solution of NaCl means there are 2.23 moles of NaCl in 1000 mL of solution.

We know that density is equal to ratio of mass to volume.

Here density of solution is 1.01 g/mL.

So mass of 1000 mL solution = ( $1.01\times 1000$ ) g = 1010 g

molar mass of NaCl = 58.44 g/mol

So mass of 2.23 moles of NaCl = ( $2.23\times 58.44$ ) g = 130.3 g

% by mass is ratio of mass of solute to mass of solution and then multiplied by 100.

Here solute is NaCl.

So % by mass of 2.23 M aqueous solution of NaCl = $\frac{130.0}{1010}\times 100$ % = 12.90%

3 0

2 years ago