Since we assume both reactants have 1 equivalent of H+ and OH- ions, we can balance the moles out. The acid of concentration x will have (x M)(0.035 L) = 0.035x moles of acid. Meanwhile, for the base: (0.432 M)(0.0246 L) = 0.0106 moles of base. Since these must be equivalent:
0.035x = 0.0106x = 0.304 M
Answer: Climax Commuity.
Explanation: (hope it’s right! Good luck!)
5.512 litres is the volume of 15.2 grams of sulphur dioxide gas at STP.
Explanation:
Data given:
mass of sulphur dioxide = 15.2 grams
conditions is at STP whech means volume = 22.4 litres
atomic mass of sulphur dioxide = 64.06 grams/mole
Number of moles is calculated as:
number of moles = 
Putting the values in the equation:
number of moles = 
= 0.23 moles
Assuming that sulphur dioxide behaves as an ideal gas, we can calculate the volume as:
When 1 mole of sulphur dioxide occupies 22.4 litres at STP
Then 0.23 moles of sulphur dioxide occupies 22.4 x 0.23
= 5.152 litres is the volume.
Answer:
0.64 L
Explanation:
Recall that
n= CV where n=m/M
Hence:
m/M= CV
m= given mass of solute =152g
M= molar mass of solute
C= concentration of solute in molL-1 = 1.5M
V= volume of solute =????
Molar mass of potassium permanganate= 158.034 g/mol
Thus;
152 g/158.034 gmol-1= 1.5M × V
V= 0.96/1.5
V= 0.64 L
C because it is and I know