this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is volume of the diluted solution to be prepared
50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL
substituting these values in the formula
1.50 M x 50.0 mL = C x 250 mL
C = 0.300 M
concentration of the final solution is A) 0.300 M
On the off chance that one of the reactants is in overabundance yet you don't know which one it is, you have to compute the hypothetical item mass for the both reactants, with a similar item, and whichever has the lower yield is the one you use to precisely depict masses/sums for the condition, since you can't have more than the non-abundance reactant can create.
Answer:
how am i suppose to put it in largest to smallest if u didnt give any numbers
Its a covalent bond for this q
Answer:
Explanation:
1 mol of ideal gas at STP occupies 22.4 (or 22.7 depending on the convention being used for STP) liters in volume. I will use 22.4 so 17.88*22.4 = 400.5 L
