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1 year ago

Which is the heaviest element produced in large stars by nuclear fusion near the end of their life cycle?

Chemistry

2 answers:

Rama09 [41]1 year ago

6 0

Iron is the heaviest element produced in large stars by nuclear fusion near the end of their life cycle .

<h3>What do you mean by the nuclear fusion ?</h3>

Nuclear fusion is the process by which two light atomic nuclei combine to form a single heavier one while releasing massive amounts of energy.

The main advantage of nuclear fusion is it is safe source for the generation of electricity.

Characteristics of nuclear fusion -:

The main characteristic of nuclear fusion is that it involves formation of a singular particle from two or more atoms.

Unlike nuclear fission, nuclear fusion is common in nature- just hard to recreate.

Nuclear fusion typically produces few radioactive particles .

Fusion works by applying a great deal of pressure to overcome .

Hence ,Iron is the heaviest element produced in large stars by nuclear fusion near the end of their life cycle .

Learn more about nuclear fusion ,here:

brainly.com/question/12701636

#SPJ2

Sholpan [36]1 year ago

4 0

Answer:

The lightest elements in the universe — hydrogen, helium, and a little lithium — were born shortly after the Big Bang.

Explanation:

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Determine the amount of heat (in kj) required to vaporize 1.55 kg of water at its boiling point. for water, δhvap

Usimov [2.4K]

Heat required to vaporize 1 mol of water from water at 100C to steam at 100C = 40.7 kJ
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Heat required to vaporize :
= 86.03 mol x 40.7 kJ

= 3501.421 kJ

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3 years ago

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2 years ago

Read 2 more answers

A gas has a volume of 62.65L at O degrees Celsius and 1 atm. At what temperature in Celsius would the volume of the gas be 78.31

Anastaziya [24]

Answer:

The volume of the gas will be 78.31 L at 1.7 °C.

Explanation:

We can find the temperature of the gas by the ideal gas law equation:

$PV = nRT$

Where:

n: is the number of moles

V: is the volume

T: is the temperature

R: is the gas constant = 0.082 L*atm/(K*mol)

From the initial we can find the number of moles:

$n = \frac{P_{1}V_{1}}{RT_{1}} = \frac{1 atm*62.65 L}{(0.082 L*atm/K*mol)*(0 + 273)K} = 2.80 moles$

Now, we can find the temperature with the final conditions:

$T_{2} = \frac{P_{2}V_{2}}{nR} = \frac{612.0 mmHg*\frac{1 atm}{760 mmHg}*78.31 L}{2.80 moles*0.082 L*atm/(K*mol)} = 274.7 K$

The temperature in Celsius is:

$T_{2} = 274.7 - 273 = 1.7 ^{\circ} C$

Therefore, the volume of the gas will be 78.31 L at 1.7 °C.

I hope it helps you!

8 0

3 years ago

"What quantitative measurement and qualitative observations can be made about the interactions of matter and energy?" can someon

kirza4 [7]

Quantitative measurements are numerical values, they involve amounts and units like measuring things. Qualitative observations appeal to the five senses, like what does the interaction look and sound like

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3 years ago

Determine the limiting reactant (lr) and the mass (in g) of nitrogen that can be formed from 50.0 g n2o4 and 45.0 g n2h4. some p

Licemer1 [7]

N2O4(l) + 2 N2H4(l) → 3 N2(g) + 4 H2O(g)
1) to calculate the limiting reactant you need to pass grams to moles.
 moles is calculated by dividing mass by molar mass

mass of N2O4: 50.0 g
molar mass of N2O4 = 92.02 g/mol
molar mass of N2H4 = 32.05 g/mol.
mass of N2H4:45.0 g

moles N2O4=50.0/92.02 g/mol= 0,54 mol of N2O4
moles N2H4= 45/32.05 g/mol= 1,40 mol of N2H4

 2)
By looking at the balanced equation, you can see that 1 mol of N2O4 needs 2 moles of N2H4 to fully react . So to react 0,54 moles of N2O4, you need 2x0,54 moles of N2H4 moles
N2H4 needed = 1,08 moles.
You have more that 1,08 moles N2H4, so this means the limiting reagent is not N2H4, it's N2O4. The molecule that has molecules that are left is never the limiting reactant.

3) 1 mol of N2O4 reacting, will produce 3 mol of N2 (look at the equation)
There are 0,54 mol of N2O4 available to react, so how many moles will produce of N2?
1 mol N2O4------------3 mol of N2
0,54 mol N2O4--------x
x=1,62 mol of N2

4) the only thing left to do is convert the moles obtained, to grams.
We use the same formula as before, moles equal to mass divided by molar mass.
$moles= \frac{grams}{molar mass}$ (molar mass of N2= 28)
1,62 mol of N2= mass/ 28
mass of N2= 45,36 grams

4 0

3 years ago