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Kruka [31]
3 years ago
8

Today, nuclear power plants rely on fission. While fusion reactions have been used in nuclear bombs, many scientists and enginee

rs hope that in the future we can use fusion to produce energy. What are some possible advantages of fusion energy over fission energy?
Chemistry
1 answer:
Zigmanuir [339]3 years ago
5 0

Explanation:

nuclear fusion yields more energy than nuclear fission and the products of the reaction are not radioactive

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Which of the elements are nonmetals?
Elis [28]
The answer is E and F.

Hope this helps!!
8 0
2 years ago
Read 2 more answers
For the reaction
s2008m [1.1K]

Answer:

0.558mole of SO₃

Explanation:

Given parameters:

Molar mass of SO₃ = 80.0632g/mol

Mass of S = 17.9g

Molar mass of S = 32.065g/mol

Number of moles of O₂ = 0.157mole

Molar mass of O₂ = 31.9988g/mol

Unknown:

Maximum amount of SO₃

Solution

  We need to write the proper reaction equation.

           2S + 3O₂ → 2SO₃

We should bear in mind that the extent of this reaction relies on the reactant that is in short supply i.e limiting reagent. Here the limiting reagent is the Sulfur, S. The oxygen gas would be in excess since it is readily availbale.

So we simply compare the molar relationship between sulfur and product formed to solve the problem:

First, find the number of moles of Sulfur, S:

   Number of moles of S = \frac{mass }{molar mass}

   Number of moles of S =  \frac{17.9 }{32.065} = 0.558mole

Now to find the maximum amount of SO₃ formed, compare the moles of reactant to the product:

       2 mole of Sulfur produced 2 mole of SO₃

   Therefore; 0.558mole of sulfur will produce 0.558mole of SO₃

5 0
3 years ago
All of the following are features of karst topography EXCEPT sinkholes. Disappearing and emerging streams. Caverns. Continuous s
Minchanka [31]
D, continuous streams.
4 0
3 years ago
2. Hydrogen gas at a temperature of 22.0°C that is confined in a 5.00L cylinder exerts a pressure of 4.20atm. If the gas is rele
Umnica [9.8K]

Answer: n∗R=22+273.15/4.2∗5n

P2=n∗R∗T2/V2=n∗R∗33.6+273.15/10

Explanation:

4 0
3 years ago
A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place
kakasveta [241]

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8  cm³

mass of gold cube =  20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

7 0
3 years ago
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