The answer is 6 moles of water will be produced.
The mass of magnesium in 
atoms is 240 g.
Answer: Option A
<u>Explanation:</u>
First, we have to convert the atoms to moles of magnesium.
We know that atoms are present in 1 mole of magnesium. So,
Thus,
Thus, 240 g of Magnesium is present in atoms.
Answer:
Oxidized and reducing agent: manganese.
Reduced and oxidizing agent: mercury.
Explanation:
Hello!
In this case, for the reaction:
We keep in mind that the species that increase the oxidation state is the oxidized one whereas the one that decrease the oxidation state is the reduced one; therefore manganese is the oxidized one as well as the reducing agent because it goes from 0 to +2 and mercury the reduced one as well as the oxidizing agent because it goes from 2+ to 0.
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Answer:
<em>Hello, Your answer will be </em><em>B) Jacob's backyard is on the north side of his house.</em>
<em>Hope That Helps!</em>
Answer:
308 g
Explanation:
Data given:
mass of Fluorine (F₂) = 225 g
amount of N₂F₄ = ?
Solution:
First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄
Reaction:
2F₂ + N₂ -----------> N₂F₄
Now look at the reaction for mole ratio
2F₂ + N₂ -----------> N₂F₄
2 mole 1 mole
So it is 2:1 mole ratio of Fluorine to N₂F₄
As we Know
molar mass of F₂ = 2(19) = 38 g/mol
molar mass of N₂F₄ = 2(14) + 4(19) =
molar mass of N₂F₄ = 28 + 76 =104 g/mol
Now convert moles to gram
2F₂ + N₂ -----------> N₂F₄
2 mole (38 g/mol) 1 mole (104 g/mol)
76 g 104 g
So,
we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.
Apply unity formula
76 g of F₂ ≅ 104 g of N₂F₄
225 g of F₂ ≅ X of N₂F₄
Do cross multiplication
X of N₂F₄ = 104 g x 225 g / 76 g
X of N₂F₄ = 308 g
So,
308 g N₂F₄ can be produced from 225 g F₂
