1. What is the Molarity of a solution that has 25.00 g of NaOH in 500.00 ml of solution. 2. You're provided a 0.50 M solution of HCl. How many moles of HCl are present in 25 ml of this solution?

3. A lab tech prepared a solution of sodium carbonate, Na2CO3, with a molarity of 0.030 M. What mass of Na2CO3 will there be in 1.00 L of this solution?

4.The following equation shows the dissociation of barium hydroxide in aqueous solution:

Ba(OH)2 --> Ba2+ + 2OH-

Determine the number of moles of hydroxide ion in 250.00 ml of a 0.20 M solution of barium hydroxide.

5. If it takes 50.00 mL of 0.50 M Ca(OH)2 to neutralize 125.00 mL of sulfuric acid, what is the concentration of the acid?

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 !

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"