How much energy is required to melt these quantities of ice? a. 56 g b. 56 mol

1 answer:

4 0

The enthalpy is the amount of energy required to change from solid to liquid.

The molar enthalpy of fusion (kJ/mol) is 6.01 kJ/mol.

We can answer B first, since solving for A would require us to find moles, which takes more time.

q = H*n
q = 6.01*56
q = 336.56 kJ

On to a.

Since we know the molar enthalpy of fusion, we need to solve for the number of moles in 56 grams of ice.

n = 56/(16.00 + 2(1.008))
n = 3.108 mol

Now we can solve.

q = H * n
q = 6.01 * 3.108
q = 18.7 kJ

Answers:
a. 18.7 kJ
b. 336.56 kJ

Let me know if you'd like me to explain anything I did here.

-T.B.

You might be interested in

Can you explain what happens to resistance when voltage is increased?

Novay_Z [31]

Resistance is directly proportional to voltage so when you increase Voltage resistance will also increase in order to keep the rate of flow of charge (current) constant

Hope this helps!

4 0

3 years ago

An atom that normally has __________ electrons in its outer (valence) shell tends not to form chemical bonds with other atoms.

goblinko [34]

8 valence shell electrons.

5 0

3 years ago

Calculate the number of carbon atoms in 25.0 grams of isopropyl alcohol (c3h8o)

allochka39001 [22]

First, let's determine the number of moles of carbon atoms by using molar mass. Then, using Avogadro's number, we can find the number of C atoms:
*1 mole of C3H8O= (12.0x3)+(1.0x8)+(16.0x1) = 60.0g/mol

25.0 grams C3H8O x (1 mole C3H8O/60.0 grams) = 0.417 mol

0.417mol C3H8O has (3 x 0.417 moles) C atoms = 1.251 moles C atoms

1.251 moles C atoms x(6.022x10^23 atoms/mol) = 9.42x10^23 C atoms.

The answer is in 3 significant figures, as that's what we have in the given, and we matched it with our rounding of the atomic masses from the periodic table.
You can do this all in one equation written left to right, just exclude the intermediate answers. Just easier to show it this way on the computer screen.

7 0

4 years ago

Read 2 more answers

Why is it that you can not feel earths movement but you can observes its effects

ss7ja [257]

The correct answer would be we have no other frame of reference besides the sky. We could actually tell we are moving because of the shifting of stars in the sky over time. But really, “A” would be your answer. We have the same momentum of the earth. It’s like being on a train at a constant speed. The only way you know you’re moving is if you look out the window or the speed changes causing you to be pushed around. Same thing with earth, our only reference is the sky

5 0

4 years ago

Look at the following data provided below:

Vlad1618 [11]

Considering the Hess's Law, the enthalpy change for the reaction is -84.4 kJ.

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.

<h3>Enthalpy change for the reaction in this case</h3>

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

2 C (graphite) + 3 H₂(g) → C₂H₆(g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: C₂H₆(g) + $\frac{7}{2}$ O₂(g) → 2 CO₂(g) + 3 H₂O(l) ; ΔH° = –1560 kJ

Equation 2: H₂(g) + $\frac{1}{2}$ O₂(g) → H₂O(l) ; ΔH° = –285.8 kJ

Equation 3: C(graphite) + O₂(g) → CO₂(g) ; ΔH° = –393.5 kJ

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.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 2 moles of C(graphite) on reactant side and it is present in third equation. In this case it is necessary to multiply it by 2 to obtain the necessary amount. 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.

Now, you need 3 moles of H₂(g) on reactant side and it is present in second equation. In this case it is necessary to multiply it by 3 to obtain the necessary amount and the variation of enthalpy also is multiplied by 3.

Finally, 1 mole of C₂H₆(g) must be a product and is present in the first equation. Since this equation has 1 mole of C₂H₆(g) on the reactant side, it is necessary to locate the C₂H₆(g) on the reactant side (invert it). When an equation is inverted, the sign of delta H also changes.

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

Equation 1: 2 CO₂(g) + 3 H₂O(l) → C₂H₆(g) + $\frac{7}{2}$ O₂(g); ΔH° = 1560 kJ

Equation 2: 3 H₂(g) + $\frac{3}{2}$ O₂(g) → 3 H₂O(l) ; ΔH° = –857.4 kJ