30) how many grams of ch3oh must be added to water to prepare 210 ml of a solution that is
1.50 m ch3oh?
a) 10.0g
b) 48.0 g
c) 3.20 g
d) 0.210 g
e) 22.08

Mass = 10.0 g

Explanation:

Given data:

Molarity = 1.50 M

Volume = 210 mL ( 0.21 L)

Grams of CH₃OH = ?

Solution:

Number of mole = molarity × volume

Number of mole = 1.50 M × 0.21 L

Number of mole = 0.315 mol

Mass of CH₃OH:

Mass = Number of moles ×  molar mass

Mass = 0.315 mol × 32.04 g/mol

Mass = 10.0 g

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.