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3 years ago

Compare fission and fusion reactions and explain the difference between their energy input and output.

2 answers:

8 0

Both the two are atomic responses that deliver vitality, however, the applications are not the same. Parting is the part of a substantial, flimsy core into two lighter cores, and the combination is the procedure where two light cores join together discharging tremendous measures of vitality.

sertanlavr [38]3 years ago

7 0

Answer :

Nuclear fusion : It is defined as a process which involves the conversion of two small nuclei to form a heavy nuclei and some amount of energy is also releases.

Nuclear fission : It is defined as a process which involves the conversion of a heavier nuclei (which is an unstable nuclei) into two or more small nuclei (which is a stable nuclei) and a large amount of energy is also releases.

The difference between the nuclear fusion and nuclear fission is that, the energy is given in the nuclear fusion reaction but the energy is releases in the nuclear fission reaction.

In nuclear fusion, a small amount of energy is released but in nuclear fission, more amount of energy is released.

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Which electron configuration represents an atom of aluminum in an excited state?

weeeeeb [17]

I’m not sure but I think it’s 2-7-4 because the numbers should still add up to 13.(hope it helped)

4 0

3 years ago

Most of the sulfur used in the United States is chemically synthesized from hydrogen sulfide gas recovered from natural gas well

Harlamova29_29 [7]

Answer: Rate of production of Sulfur dioxide is 0.15kg/s to 2 significant digits

Note: The question is incomplete. The complete question is as follows:

Most of the used in the United States is chemically synthesized from hydrogen sulfide gas recovered from natural gas wells. In the first step of this synthesis, called the Claus process, hydrogen sulfide gas is reacted with dioxygen gas to produce gaseous sulfur dioxide and water Suppose a chemical engineer studying a new catalyst for the Claus reaction finds that 198. liters per second of dioxygen are consumed when the reaction is run at 186. °C and 0.69 atm. Calculate the rate at which sulfur dioxide is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.

Explanation:

The balanced equation of the reaction between hydrogen sulfide gas and dioxygen gas to produce sulfur dioxide and water is as follows:

2H₂S(g) + 3O₂(g) ---> 2SO₂(g) + 2H₂O(g)

From the equation above, 3 moles of dioxide yields 2 moles of sulfur dioxide.

Using the ideal gas equation to determine the number of moles of dioxygen present in 198 litres of the gas at 186. °C and 0.69 atm;

PV = nRT,

where P = 0.69atm, V = 198Litres, R (molar gas constant) = 0.082atmLK⁻¹mol⁻¹, T = 186. °C = 459K

n = PV/RT

n = 0.69*198*/(0.082*459)

n = 3.63moles of dioxygen

Therefore, 3.63 moles of dioxygen are consumed per second

Using the mole ratio from the equation of reaction,

number of moles of sulfur dioxide produced will be 3.63moles * 2/3 = 2.42moles

Therefore, the number of moles of sulfur dioxide consumed is 2.42 moles per second.

Converting to kilograms per second,

1 mole of sulfur dioxide weighs 64g (molar mass of sulfur dioxide)

2.42 moles weighs 2.42*64g =154.88g or 0.15488Kg which is approximately 0.15Kg

Therefore, rate of production of Sulfur dioxide is 0.15kg/s to 2 significant digits

8 0

3 years ago

2 H2(g) + O2(g) → 2 H2O(g)

alexgriva [62]

Answer:

77 L of water can be made.

Explanation:

Molar mass of $O_{2}$ = 32 g/mol

So, 55 g of $O_{2}$ = $\frac{55}{32}$ mol of $O_{2}$ = 1.72 mol of $O_{2}$

As hydrogen is present in excess amount therefore $O_{2}$ is the limiting reagent.

According to balanced equation, 1 mol of $O_{2}$ produces 2 mol of $H_{2}O$ .

So, 1.72 mol of $O_{2}$ produce $(2\times 1.72)$ mol of $H_{2}O$ or 3.44 mol of $H_{2}O$ .

Let's assume $H_{2}O$ gas behaves ideally at STP.

Then, $P_{H_{2}O}.V_{H_{2}O}=n_{H_{2}O}.R.T$ , where P, V, n, R and T represents pressure, volume, no. of moles, gas constant and temperature in kelvin scale respectively.

At STP, pressure is 1 atm and T is 273 K.

Here, $n_{H_{2}O}$ = 3.44 mol and R = 0.0821 L.atm/(mol.K)

So, $(1atm)\times V_{H_{2}O}=(3.44mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)$

$\Rightarrow$ $V_{H_{2}O}=77L$

Option (b) is correct.

5 0

3 years ago

What is a mammal and why are they called mammals

taurus [48]

Answer:

mammals are the animal who lay eggs

7 0

3 years ago

Read 2 more answers

Calculate the number of moles of H, in 11 L of gas at STP.

kramer

Explanation:

1 mole of a gas has 22.4 L of volume at STP. Therefore, 0.5 mol of H2 has 0.5 x 6.022 x 1023 = 3.011 x 1023 molecules. Hence, the number of moles, molecules and atoms present in 11.2 liters of H2 at STP are 0.5 , 3.011 x 1023 and 6.022 x 1023 respectively.

3 0

3 years ago