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3 years ago

Which describes energy changes during nuclear fission?

Chemistry

1 answer:

yarga [219]3 years ago

3 0

Answer:

A very large amount of energy

is produced from a very small mass.

Background Info:

Nuclear fission is the process of splitting apart nuclei (usually large nuclei). When large nuclei, such as uranium-235, fissions, energy is released. So much energy is released that there is a measurable decrease in mass, from the mass-energy equivalence. This means that some of the mass is converted to energy.

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The standard enthalpy change for the reaction of SO3(g) with H2O(l) to yield H2SO4(aq) is ΔH∘ = -227.8 kJ . Use the following in

olga nikolaevna [1]

Answer:

-909.3KJ/mole

Explanation:

The heat of reaction is accessible from the heat of formation of reactants and products using the formula below:

ΔH = Σ ΔHf products - Σ ΔHf reactants

Before we proceed, it is important to know that the enthalpy of formation of element is zero ,be it a single element or a molecule of an element.

From the reaction for the formation of sulphuric acid, we know we need to know the heat of formation of sulphur (vi) oxide and water. The examiner is quite generous and have us for water already.

Now we need to calculate for sulphur (vi) oxide. This is calculated as follows:

We first calculate for sulphur(iv)oxide. This can be obtained from the reaction between sulphur and oxygen. The calculation goes thus:

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole suphur(iv) oxide × x] - [ (1 mole of elemental sulphur × 0) + (1 mole of elemental oxygen × 0]

We were already told this is equal to -296.8KJ. Hence the heat of formation of sulphur(iv) oxide is -296.8KJ.

We then proceed to the second stage.

Now, here we have 1 mole sulphur (iv) oxide reacting with 0.5 mole oxygen molecule.

We go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphur (vi) oxide × y] - [ (1 mole of sulphur (iv) oxide × -296.8) + (0.5 mole of oxygen × 0)].

We already know that the ΔH here equals -98.9KJ.

Hence, -98.9 = y + 296.8

y = -296.8KJ - 98.9KJ = -395.7KJ

We now proceed to the final part of the calculation which ironically comes first in the series of sentences.

Now, we want to calculate the standard heat of formation for sulphuric acid. From the reaction, we can see that one mole of sulphur (vi) oxide, reacted with one mole of water to yield one mole of sulphuric acid.

Mathematically, we go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphuric acid × z] - [( 1 mole of sulphur vi oxide × -395.7) + ( 1 mole of water × -285.8)].

Now, we know that the ΔH for this particular reaction is -227.8KJ

We then proceed to to open the bracket.

-227.8 = z - (-395.7 - 285.8)

-227.8 = z - ( -681.5)

-227.8 = z + 681.5

z = -227.8-681.5 = -909.3KJ

Hence, ΔH∘f for sulphuric acid is -909.3KJ/mol

6 0

3 years ago

If 1 kilowatt-hour (kWh) = 3.60 × 10^6 J, which of the following conversion factors should be used to convert 56.7 kWh to J?

Nitella [24]

Answer:

c option is correct

Explanation:

May this help you!!!!!!

3 0

3 years ago

A container is filled with a mixture of helium, Xenon, and neon gases. The total pressure inside the container is 1,788 mmHg. If

dmitriy555 [2]

Answer: The partial pressure of Xenon in a container filled with a mixture of helium, Xenon, and neon gases and a total pressure inside the container is 1,788 mmHg with the pressure of the helium gas as 760 mmHg, the pressure of the neon gas as 200 mmHg is 828 mmHg

Explanation:

NOTE :

Dalton's law of partial pressures states that in a mixture of gases the pressure exerted by each gas is the same as that which it would exert if it alone occupied the container.

How to determine the partial pressure

Total pressure = Partial pressure of Neon + Partial pressure of Xenon + Partial pressure of helium

Partial pressure of Xenon = ? mmHg

Partial pressure of Helium = 760 mmHg

Partial pressure of Neon = 200 mmHg

Total pressure of gases = 1, 788 mmHg

Partial pressure of Xenon = Total pressure - Partial pressure of Helium - Partial pressure of Neon

Partial pressure of Xenon = 1, 788 - 760 -200

Partial pressure of Xenon = 1,788 - 960

= 828 mmHg

Thus, the partial pressure of Xenon is 828 mmHg

6 0

2 years ago

What is the molality of a solution of water and kcl if the freezing point of the solution is –3mc030-1.jpgc?

Natasha_Volkova [10]

We will use the expression for freezing point depression ∆Tf
∆Tf = i Kf m
Since we know that the freezing point of water is 0 degree Celsius, temperature change ∆Tf is
∆Tf = 0C - (-3°C) = 3°C
and the van't Hoff Factor i is approximately equal to 2 since one molecule of KCl in aqueous solution will produce one K+ ion and one Cl- ion:
KCl → K+ + Cl-
Therefore, the molality m of the solution can be calculated as
3 = 2 * 1.86 * m
m = 3 / (2 * 1.86)
m = 0.80 molal

4 0

4 years ago

What pressure would a gas exert at absolute zero?Explain

valina [46]

The term absolute zero is given because the volume and the temperature comes to 0. So according to Charles law,
At const K temp, Volume is proportional to Temperature. So the volume is also 0. P = nRT/V. T=0 and P=0

7 0

3 years ago