The answer is (2) release a large amount of energy. Nuclear fission form light nuclides from heavy nuclides. While nuclear fusion form heavy nuclides from light nuclides.
P₄0₆
+ 20₂
⇒ P₄0₁₀
Explanation:
The overall equation for the reaction that produces P₄0₁₀ is :
P₄0₆ + 20₂ ⇒ P₄0₁₀
Now let us derive this equation:
Given equations:
P₄ + 30₂ ⇒ P₄0₆ equation 1;
P₄ + 50₂ ⇒ P₄0₁₀ equation 2;
To get the overall combined equation, the equation 1 must be reversed and added to equation 2:
P₄0₆ ⇒ P₄ + 30₂ equation 3
+
equation 2:
P₄ + 50₂ + P₄0₆ ⇒ P₄0₁₀ + P₄ + 30₂
cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;
P₄0₆ + 20₂ ⇒ P₄0₁₀
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You have to find the stoichiometric ratio between AlCl₃ and BaCl₂. The common element between them is Cl. So, the ratio of Cl in BaCl₂ to AlCl₃ is 2/3. The molar mass of AlCl₃ is 133.34 g/mol. The solution is as follows:
Mass of AlCl₃ = (6 mol BaCl₂)(2 mol Cl/1 mol BaCl₂)(1 mol AlCl₃/3 mol Cl)(133.34 g/mol) = 533.36 g AlCl₃
Answer: This is because the number of shell increases .
Explanation: On moving from left to write on the periodic table the reactivity of non metals increases because number of shells increases and the force with which the nucleus hold electrons decreases.
Answer:
14 mol O₂
Explanation:
The reaction between CO and O₂ is the following:
CO + O₂ → CO₂
We balance the equation with a coefficient 2 in CO and CO₂ to obtain the same number of O atoms:
2CO + O₂ → 2CO₂
As we can see from the balanced equation, 1 mol of O₂ is required to react with 2 moles of CO. Thus, the conversion factor is 1 mol of O₂/2 mol CO. We multiply the moles of CO by the conversion factor to calculate the moles of O₂ that are required:
28 mol CO x 1 mol of O₂/2 mol CO = 14 mol O₂
