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3 years ago

A recreational runner has an average speed of 5.9 mph. How many km does she cover in 98 min?

Chemistry

1 answer:

mylen [45]3 years ago

7 0

Given

Speed = 5.9 miles / hour

1 mile = 1.609 km

1 hour = 60 min

Time given = 98 min

Convert miles to km, hours to min and calculate the distance for 98 min:

pic added above

Please cooperate with answer as am a new to brainly so I don't know much about it.

Thank you.

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Using the equations

Anna [14]

Considering the Hess's Law, the enthalpy change for the reaction is 221.8 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol

Equation 2: C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol

Equation 3: 2 C(s) + 2 H₂ (g) → C₂H₄ (g) ∆H° = -97.6 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

<h3 /><h3>FIRST STEP</h3>

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C₂H₄ (g) on reactant side and it is present in first equation. Since this equation has one mole of C₂H₄ (g) on the product side, it is necessary to locate it on the reactant side (invert it).

When an equation is inverted, the sign of ΔH° also changes.

<h3>SECOND STEP</h3>

Now, you need 2 moles of CF₄ (g) on the product side. The second equation has 1 mole of CF₄ (g) on the product side, so it is necessary to multiply it by 2 to obtain 2 moles of CF₄ (g).

Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

<h3>THIRD STEP</h3>

Finally, you need 4 moles of HF (g) on the product side. The first equation has 2 moles of HF (g) on the product side, so it is necessary to multiply it by 2 to obtain 4 moles of the compound.

Since the equation is multiply by 2, the variation of enthalpy also is multiplied by 2.

<h3>SUMMARY</h3>

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 H₂ (g) + 2 F₂ (g) → 4 HF (g) ∆H° = -158.4 kJ/mol

Equation 2: 2 C (s) + 4 F₂ (g) → 2 CF₄ (g) ∆H° = 282.6 kJ/mol

Equation 3: C₂H₄ (g) → 2 C(s) + 2 H₂ (g) ∆H° = 97.6 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g) ΔH°= 221.8 kJ/mol

Finally, the enthalpy change for the reaction is 221.8 kJ/mol.

Learn more about molar enthalpy:

7 0

2 years ago

I) Briefly discuss the strengths and weaknesses of the four spectroscopy techniques listed below. Include in your answer the spe

klemol [59]

delete please .....................................

4 0

3 years ago

COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at

mestny [16]

<u>Answer:</u> The value of $K_p$ for the reaction at 690 K is 0.05

<u>Explanation:</u>

We are given:

Initial pressure of $COCl_2$ = 1.0 atm

Total pressure at equilibrium = 1.2 atm

The chemical equation for the decomposition of phosgene follows:

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

Initial: 1 - -

At eqllm: 1-x x x

We are given:

Total pressure at equilibrium = [(1 - x) + x+ x]

So, the equation becomes:

$[(1 - x) + x+ x]=1.2\\\\x=0.2atm$

The expression for $K_p$ for above equation follows:

$K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}$

$p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm$

Putting values in above equation, we get:

$K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05$

Hence, the value of $K_p$ for the reaction at 690 K is 0.05

3 0

3 years ago

a chemist wishes to mix some pure acid with some water to produce 16L of a solution that is 30% acid how much pure acid and how

Aloiza [94]

<u>Answer:</u> The volume of acid and water that must be mixed will be 4.8 L and 11.2 L

<u>Explanation:</u>

We are given:

Volume of mixture = 16 L

Percent of acid present = 30 %

Calculating the percentage of acid present in the mixture:

$\Rightarrow 16\times \frac{30}{100}=4.8L$

The mixture is made entirely of acid and water.

Volume of acid in the mixture = 4.8 L

Volume of water in the mixture = 16 - 4.8 = 11.2 L

Hence, the volume of acid and water that must be mixed will be 4.8 L and 11.2 L

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3 years ago

What products are always made when any carbon-based fuel

Elodia [21]

Answer:
d) carbon dioxide and water

8 0

3 years ago