at STP volume occupied by 1 mol of gas= 22.4 L
given volume = 0.224 m3= 0.000224 L
so no. of moles of gas= 0.000224/22.4
= 10-5 moles
1 mole = 6.02* 1023 particles
10-5 moles = 6.02 * 1018 particle.
2).
pH=12
pOH = 14-12= 2
[OH-] = 0.01 M
mmoles of OH- = 0.01*100 = 1 mmol
total volume = 100+900 mL = 1000mL
concentration of OH-= 1/1000 = 10-3 M
pOH = - log[OH-]
pOH= 3
pH= 14-3=11
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