I think A.weathering, is the best possible answer
Answer:
The answer to your question is MgSO₄ 5H₂O
Explanation:
Data
mass of MgSO₄ = 2.86 g
mass of H₂O = 2.14 g (5 - 2.86)
Process
1.- Calculate the molecular mass of the compounds
MgSO₄ = 24 + 32 + (16 x 4) = 120
H₂O = 16 + 2 = 18
2.- Convert the grams obtain to moles
120 g of MgSO₄ --------------- 1 mol
2.8 g ---------------- x
x = (2.8 x 1)/120
x = 0.024 moles
18 g of H₂O --------------------- 1 mol
2.14 g -------------------- x
x = (2.14 x 1)/18
x = 0.119
3.- Divide by the lowest number of moles
MgSO₄ = 0.024/0.024 = 1
H₂O = 0.119/ 0.024 = 5
4.- Write the molecular formula
MgSO₄5H₂O
Answer:
The number of atoms in the outermost shell
Explanation:
For example, the electron shells in the alkali metals contain the following numbers of electrons:
Li: 2, 1
Na: 2, 8, 1
K: 2, 8, 8, 1
They all have one electron in their outermost shell, and they have similar chemical properties.
I believe it’s the last option
Answer:
5 moles of NO₂ will remain after the reaction is complete
Explanation:
We state the reaction:
3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g)
3 moles of nitric oxide can react with 1 mol of water. Ratio is 3:1, so we make this rule of three:
If 3 moles of nitric oxide need 1 mol of water to react
Then, 26 moles of NO₂ may need (26 .1) / 3 = 8.67 moles of H₂O
We have 7 moles of water but we need 8.67 moles, so water is the limiting reactant because we do not have enough. In conclusion, the oxide is the reagent in excess. We can verify:
1 mol of water needs 3 moles of oxide to react
Therefore, 7 moles of water will need (7 .3)/1 = 21 moles of oxide
We have 26 moles of NO₂ and we need 21, so we still have oxide after the reaction is complete. We will have (26-21) = 5 moles of oxide that remains
