It’s is m2 +11m-11
I believe that’s right
Answer:
The empirical formula is ZnO2
Explanation:
What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?
Step 1: Data given
Suppose the compound has a mass of 100.0 grams
A compound contains:
67.1 % Zinc = 67.1 grams
100 - 67.1 = 32.9 % oxygen = 32.9 grams
Molar mass of Zinc = 65.38 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate moles of Zinc
Suppose the compound is 100 grams
Moles Zn = 67. 10 grams / 65.38 g/mol
Moles Zn = 1.026 moles
Step 3: Calculate moles of O
Moles O = 32.90 grams / 16.00 g/mol
Moles O = 2.056 moles
Step 4: Calculate mol ratio
We divide by the smallest amount of moles
Zn: 1.026/1.026 = 1
O: 2.056/1.026 = 2
The empirical formula is ZnO2
To control this we can calculate the % Zinc for 1 mol
65.38 / (65.38+2*16) = 0.67.1 = 67.2 %
Answer:
CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)
Explanation:
The balanced reaction between Na2CO3 and HCl is given as;
Na₂CO₃ (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO₂ (g) + H₂O (l)
The next step is o express the species as ions.
The complete ionic equation for the above reaction would be;
2Na⁺(aq) + CO₃²⁻(aq) + 2H⁺(aq) + 2Cl⁻(aq) → Na⁺(aq) + Cl⁻(aq) + CO₂ (g) + H₂O (l)
The next step is to cancel out the spectator ion ions; that is the ions that appear in both the reactant and product side unchanged.
The spectator ions are; Na⁺ and Cl⁻
The net ionic equation is given as;
CO₃²⁻(aq) + 2H⁺(aq) → CO₂ (g) + H₂O (l)
0.5 litres contain 1.1 moles
therefore 1 litre will have= 1*1.1/0.5
molarity=2.2M
