Answer:
So the answer would be 10 moles
Explanation:
1) Start with the molecular formula for water: 
2) If there are 10 moles of water use a mole ratio to calculate the moles of oxygen it would produce.
(This question is... interesting... since they chose an element that is diatomic in free state so It could TECHNICALLY be two answers, moles of O or moles of 
)
The mole ratio is 1 moles of 
to 1 moles of O. This is because the coefficient for oxygen in water is simple 1, so the ratio is 1:1.
3) that means if 10 moles of water decompose, they decompose into 10 moles of 
and 10 moles of O.
Extra:
About what I was saying before about the question being slightly interesting:
10 moles of pure oxygen is produced but free state oxygen exists as so it could possibly be 10 OR 5! However, notice it says elements. This leads me to believe the answer is 10 (monatomic oxygen) instead of 5 (free state/diatomic oxygen).
I hope this helps!
Answer:
2 mole of Sodium hydroxide reacts with 1 mole of Sulfuric acid
Explanation:
Write down the equation in the beginning with reactants and products:
NaOH + H₂SO₄ → Na₂SO₄ + H₂0
Now try to balance it. Try with Na first:
2NaOH + H₂SO₄ → Na₂SO₄ + H₂0
Na atoms are balanced. There are 6 Oxygen atoms on the right and 5 on the left. Balance by increasing the H₂O moles:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0
Check if H atoms are also balanced. They are. That means our final reaction is:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0
2 Moles of NaOH reacts with 1 mole of H₂SO₄
Answer:
2.30 × 10⁻⁶ M
Explanation:
Step 1: Given data
Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M
Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³
Step 2: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂
We will use the following expression.
Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²
[OH⁻] = 2.30 × 10⁻⁶ M
Answer: Pt(Cl)2(NH3)2
Explanation:
In the formation of the complex, the oxidation number of platinum is plus two (+2) and two chloride ions cancel it out by their oxidation number of -1 each. Hence the complex has an overall charge of zero. It is thus neutral with no charge attached to its formula.
