Answer:
Explanation:
Nuclear fusion of hydrogen into helium occurs in the
1 answer:
There are types of nuclear reaction: nuclear fusion and nuclear fission. The difference is that fusion is a combination of two elements while fission is the breaking up of the subatomic particles of an element creating a new element. The limiting element to this is Iron. Iron-26 is the most stable element. As a result, elements lighter than Fe-26 are generally fusible. This includes hydrogen and helium.
This reaction is common in the stars, most especially the Sun. The energy of the Sun comes from its abundant hydrogen composition which becomes fusible into Helium. This occurs at a temperature of 14 million Kelvin. The nuclear reaction is a not a one-way step process as shown in the picture.
This reaction is common in the stars, most especially the Sun. The energy of the Sun comes from its abundant hydrogen composition which becomes fusible into Helium. This occurs at a temperature of 14 million Kelvin. The nuclear reaction is a not a one-way step process as shown in the picture.
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Answer:
Explanation:
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In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:
Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:
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Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂: