# Chemistry help 10 points for answer!!!!!!!!!!!!!!!?

Hey guys i have a chemistry test coming up tommorow and i have a couple questions on my study guide that i need help with. I need these as soon as possible so the person who answers will get 10 points!

Write the chemical formula of the neutral compound formed from the ions below

1. Cu2+ ions and NO3– ions

2. Ag+ ions and CrO42– ions

3. There are two possible methods for recovering copper from a solution of copper (II) sulfate, one uses precipitation and the other uses redox.

(

a) Copper(II) ions can be precipitated as copper(II) carbonate.

CuSO4(aq) + Na2CO3(aq) → CuCO3(aq) + Na2SO4(aq)

How many mL of 0.139 M Na2CO3 would be needed to precipitate all of the copper ions in 27.5 mL of 0.170 M CuSO4?

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• Anonymous
9 years ago

HERE it is:

Write the chemical formula of the neutral compound formed from the ions below

1. Cu2+ ions and NO3– ions

[Cu]2+ + [NO3]- > Cu(NO3)2

2. Ag+ ions and CrO42– ions

[Ag]+ + [CrO4]2- > Ag2CrO4

3. There are two possible methods for recovering copper from a solution of copper (II) sulfate, one uses precipitation and the other uses redox. a) Copper(II) ions can be precipitated as copper(II) carbonate.

How many mL of 0.139 M Na2CO3 would be needed to precipitate all of the copper ions in 27.5 mL of 0.170 M CuSO4?

Step 1: Write out the molecular formula:

CuSO4(aq) + Na2CO3(aq) → CuCO3(aq) + Na2SO4(aq)

You can see that it is a 1:1 ratio.

Step 2: Find molar masses: Optional (only if you need to find grams for any part of your answer):

Cu = 1(63.546)

S = 1(32.066)

O = 4(15.999)

CuSO4 = 159.608 g/mol

Na = 2(22.990)

C = 1(14.007)

O = 3(15.999)

Na2CO3 = 107.9848 g/mol

Step 2.5 :Convert mL to L:

(27.5.0 mL / 1000) = 0.0275 L

Step 3: Find moles:

27.5 mL of 0.170 M CuSO4

0.170 M CuSO4 = (x mol CuSO4 / 0.0275 L CuSO4

(0.170 M CuSO4 * 0.0275 L CuSO4) = x CuSO4

x mol CuSO4 = 0.004675 mol CuSO4

x mol CuSO4 = x mol Na2CO3

x mol Na2CO3 = 0.004675 mol Na2CO3

Step 4: Find volume:

0.139 M Na2CO3 = (x mol Na2CO3 / x L Na2CO3)

0.139 M Na2CO3 = (0.004675 mol Na2CO3 / x L Na2CO3)

x L Na2CO3 = (0.004675 mol Na2CO3 / 0.139 M Na2CO3)

x L Na2CO3 = 0.0336 L = 33.6 mL

If you know what you are doing and you make sure to write it out step-by-step it is not too hard.

Hope I helped! :)

contact me at olijake999@gmail.com for any future problems. Thank you!

Source(s): Honors Chemistry