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3 years ago

Calculate the amount of heat required to melt 2108 g of water. The enthalpy of fusion of water is ΔHfus=6.010 kJ/mol.

1 answer:

6 0

since the unit for the heat of fusion is kJ/mol, you're going to have to convert the grams into moles in order to cancel out the unit. After that, you can solve like normal.

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Fofrffycfcfyguffucggr

ira [324]

Answer:

In his alien language, he said ¨ Hellooooo human¨

Explanation:

4 0

3 years ago

Some nitrogen is held in a 2.00-L tank at a pressure of 3.00 atm. The tank is connected to a 5.00-L tank that is completely empt

leva [86]

Answer:

0.857 atm

Explanation:

The nitrogen stops owing when it fulfills both of the tanks (the gas molecules intend to fulfill all the space they are). So the tanks will have the same pressure, and the final volume will be the volume of the two tanks.

For Boyle's law:

P1*V1 = P2*V2

Where P1 is the initial pressure (3.00 atm), V1 is the initial volume(2.00 L), P2 is the final pressure, and V2 is the final volume (2.00 + 5.00 = 7.00 L).

3.00*2.00 = P2*7.00

7.00P2 = 6.00

P2 = 0.857 atm

7 0

3 years ago

A solution contains 72.0 g of HCI and 468 g of (C6H6). what is the mole fraction of benzene

erica [24]

Molar mass (MM) of benzene C6H6
C = 6 * 12 = 72u
H = 6 * 1 = 6u
MM C6H6 = 72 + 6 = 78 g / mol
Benzene - Molar Mass = 78 g --------- 1 mol
Of A Mix has 468 g -------------- x

78x = 468
X = 468/78
X = 6 moles

Molar mass (MM) of Hydrochloric Acid HCl H = 1 * 1 = 1u
CI = 1 * 35 = 35u
MM HCl = 1 + 35 = 36 g / mol

Hydrochloric Acid - Molar Mass = 36 g ---------- 1 mol
Of A Mix has 72 g ------------ y

36y = 72
Y = 72/36
Y = 2 moles

Thus, a mixture has a total of 8 moles (6 mol + 2 mol).
Dividing One Mole Amount of Each Substance by the Number of Total Mole Amounts, Then we will obtain a Molar Fraction of Each:
Molar fraction make benzene = (6/8) simplify 2 = 3/4
Molar Fraction to make Hydrochloric Acid = (2/8) = simplify 2 = 1/4

Note:. The sum of the molar fractions of the always give goes 1, we have: 3/4 + 1/4 = 1

ANSWER:
 $\boxed{\frac{3}{4}}$

4 0

3 years ago

From this figure, why are there only two possible values for the spin quantum number?

V125BC [204]

Answer:

bcoz there is only two magnetic file orienation is possible

Explanation:

7 0

3 years ago

Identify all of the following three changes that would shift the equilibrium of the Haber [HAH-bur] process to the right: a) Add

Vladimir [108]

Answer:

a).Adding hydrogen gas : Forward direction

b) Adding an inert gas

At constant pressure : Backward direction
At constant Volume : No change

c) Increasing the volume of the container:Backward direction

Explanation:

According to Le-Chatelier's principle = Any disturbance is created in the equilibrium , it will shift the equilibrium in such a way to reduce disturbance.

This disturbance can be : Change in concentration , pressure, temperature , Addition of gas etc.

Temperature : when temperature is increased for endothermic reaction . The equilibrium shift in forward. For exothermic reaction , on increasing temperature ,equilibrium shift backward.

Pressure : On increasing pressure the equilibrium shifts towards less number of gaseous molecules.

Volume : Since volume varies inversely with pressure . It's effect is opposite to that produced by pressure change i.e On increasing Volume the equilibrium shifts towards more number of gaseous molecules.

On Addition of Inert gas : The equilibrium shift toward the side where more gaseous molecules are present.

Concentration : On increasing concentration , the equilibrium will shift such that the substance added get removed .

Haber Process :

$N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)$

This is exothermic reaction (H = negative)

Molecules of gases are more on left side(reactants) than right side(product)

a) Adding hydrogen gas :

According to Le-Chatelier's principle ,The equilibrium will shift in a direction where the Hydrogen gas is consumed (removed)

Hence Equilibrium shift in forward direction

b) Adding an inert gas : There are two situation

If inert gas is added at constant pressure: Total Volume is increased, so equilibrium will shift in direction where there is increase in number of molecule(Le-Chatelier's).Here, equilibrium shift in backward direction
If inert gas is added at constant Volume :No change in equilibrium.This because if inert gas is added pressure will increase but there is no change in moles per unit volume(constant) of gases.

c) Increasing the volume of the container:

If total Volume is increased, the number of molecules per unit volume decrease.In order to maintain the equilibrium constant , the equilibrium shift in direction where there is increase in number of molecule(Le-Chatelier's).

In this case Backward direction.

3 0

3 years ago