A chemist wants to make 150 mL of a 1.26 M solution of carbonic acid to be used in the lab. How many grams of H2CO3 does the chemist need to make this solution? I converted mL to .150L, multiplied .150*1.26= =0.189 to get mol. H2CO3=62.028 so I multiplied this by 0.189 to get grams.. I got 11.722 but it's incorrect

I expect you are reporting too many significant figures. You're allowed just three; therefore, I would round the answer to 11.7 g. By the way, you will have some trouble making 1.26 M solution of H2CO3. A saturated solution isn't that strong.

I put 11.7 and it still marked me wrong :(

Your calculations are correct. To convert mL to L, you multiply by 0.001, so 150 mL becomes 0.150 L. Then, you multiply 0.150 L with the molarity (1.26 M) to calculate the number of moles:

0.150 L x 1.26 M = 0.189 mol

The molar mass of H2CO3 (carbonic acid) is approximately 62.03 g/mol. To convert moles to grams, you multiply by the molar mass:

0.189 mol x 62.03 g/mol = 11.72067 g

Rounding to the appropriate number of significant figures, the chemist would need approximately 11.72 grams of H2CO3 to make the solution. So your answer of 11.722 grams is correct.

To find the grams of H2CO3 needed to make a 1.26 M solution in 150 mL, you need to follow these steps:

1. Convert mL to liters: 150 mL is equivalent to 0.150 L.

2. Use the formula for molarity: Molarity (M) = moles (mol) / volume (L).

Rearrange the formula to solve for moles: Moles (mol) = Molarity (M) * Volume (L).

Plugging in the given values: Moles (mol) = 1.26 M * 0.150 L = 0.189 mol.

3. Determine the molar mass of H2CO3 (carbonic acid). H2CO3 consists of two hydrogen atoms (H), one carbon atom (C), and three oxygen atoms (O):

Molar mass (H2CO3) = (2 * atomic mass of H) + atomic mass of C + (3 * atomic mass of O)
= (2 * 1.01 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol)
= 62.03 g/mol.

4. Calculate the grams of H2CO3 needed:

Grams = Moles * Molar mass
= 0.189 mol * 62.03 g/mol
= 11.73 g.

Therefore, to make a 1.26 M solution of carbonic acid with a volume of 150 mL, the chemist needs 11.73 grams of H2CO3.