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4 years ago

Which of the following reactant pairs will not react together to produce a reaction?

1 answer:

5 0

I think the correct answer from the choices listed above is option D. the substances cobalt and barium chloride will not react since cobalt os less active than barium so it does not have enough energy to displace barium in the compound. Hope this answers the question.

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Consider the following reaction:C2H4(g) + F2(g) -----------> C2H4F2(g) Delta H = -549 kJEstimate the carbon-fluorine bond ene

frutty [35]

Answer:

Bond energy of carbon-fluorine bond is 485 kJ/mol

Explanation:

Enthalpy change for a reaction, is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

So, $-549kJ=(1mol\times 614kJ/mo)+(4mol\times E_{C-H})+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(4mol\times E_{C-H})-(2mol\times E_{C-F})$

or, $-549kJ=(1mol\times 614kJ/mo)+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(2mol\times E_{C-F})$

or, $E_{C-F}=485kJ/mol$

So bond energy of carbon-fluorine bond is 485 kJ/mol

8 0

3 years ago

In the combined gas law, if the pressure in the system is increased, what happens to the volume if temperature remains constant?

ANEK [815]

Boyle's law states that, at a constant temperature, the volume of a given mass of gas varies inversely with pressure. ... Thus Charles's law states that at a constant pressure, the volume of a given mass of gas is directly proportional to its (absolute)temperature.

7 0

3 years ago

Read 2 more answers

The transfer of energy that occurs when a force is applied over a distance is

nika2105 [10]

The transfer of energy that occurs when a force is applied over a distance is WORK.

Hope this helps!

3 0

3 years ago

Using the following thermochemical data: 2Y(s) + 6HF(g) → 2YF3(s) + 3H2(g) ΔH° = –1811.0 kJ/mol 2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2

Luba_88 [7]

Answer:

ΔH° = 182.4 kJ/mol

Explanation:

The ΔH wanted is for the reaction :

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g)

This is a Hess Law problem where e will have to algebraically manipulate the first and second equations , add them together, and arrive at the desired equation above.

Notice if we reverse the first equation and divide it by 2 and add to the the second only divided by two, we will arrive to the desired equation:

2YF3(s) + 3H2(g) → 2Y(s) + 6HF(g) ΔH° = 1811.0 kJ/mol (change sign)

dividing by two :

YF3(s) + 3/2H2(g) → Y(s) + 3HF(g) ΔH° = 905.5 kJ/mol Eq 1

2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2(g) ΔH° = –1446.2 kJ/mol

dividing this one by two,

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

Now adding 1 and 2

YF3(s) + 3/2H2(g) → Y(s) + 3HF(g) ΔH° = 905.5 kJ/mol Eq 1

Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2

________________________________________________________

YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g). ΔH° = 905.5 + (-723.1) kJ/mol

ΔH° = 182.4 kJ/mol

Notice how the Y(s) and H2 cancel nicely and the coefficients are the right ones.

8 0

4 years ago

When you mix the primary colors of light, you produce what color?

gayaneshka [121]

If you mix all of the primary colors, you get White light

3 0

3 years ago

Read 2 more answers