Question
A student determines the cobalt(II) content of a solution by first precipitating it as cobalt(II) hydroxide, and then decomposing the hydroxide to cobalt(II) oxide by heating. How many grams of cobalt(II) oxide should the student obtain if her solution contains 46.0 mL of 0.561 M cobalt(II) nitrate?
Answers
Co(NO3)2 ==> Co(OH)2 --> CoO
mols Co(NO3)2 = M x L = ?
mols CoO = mols Co(OH)2 = mols Co(NO3)2
grams CoO = mols x molar masss
mols Co(NO3)2 = M x L = ?
mols CoO = mols Co(OH)2 = mols Co(NO3)2
grams CoO = mols x molar masss
how many grams is that?
sorry i am stuck on many questions that's why used diff names. i need help and be grateful if u can help but most questions I don't know how to work it out.
What is confusing about this? And I suggest you pick any name and stick to it. It helps us help you if you do that.
For this problem all you need is two steps. M x L = mols and mol x molar mass = grams.
For this problem all you need is two steps. M x L = mols and mol x molar mass = grams.
i don't know much about chemitry that's why its confusing
I don't know how to calculate them
I don't know how to calculate them
Step 1. Pick out M x L. That's in the problem. M is 0.561; L is 0.046 L.
M x L = ? you can do on your calculator.
Step 2. grams = mols x molar mass
You have mols from the calculator. Multiply that by the molar mass of CoO. You can calculate that from the periodic table. Co = about 58.93 and O is 16 so 58.93 + 16 = ?
M x L = ? you can do on your calculator.
Step 2. grams = mols x molar mass
You have mols from the calculator. Multiply that by the molar mass of CoO. You can calculate that from the periodic table. Co = about 58.93 and O is 16 so 58.93 + 16 = ?
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