It is supposed to be mol/dm-3
Actually, 1 dm cube is the same as 1 litre.
Therefore, there is no need of conversion. Both are the same.
H3O= 0.9 mol/dm3 OH=1.2 mol/dm3
right i dont know this 4 sure but because u want a 0.1 mol/dm3 and u only need 100cm3 u will need 0.01mols of copper sulfate to dilute in 100cm3. soo now u have a solution that is 0.01mols per 100cm3 or 0.1 mols per 1000cm3 (dm3)
98 g per dm3 would be 1 molar - so 49 g would be 0.5 molar , so 4.9 g is 0.05 mol per dm3 or 0.05 mol.dm-3
At STP 1 mole of every gas has the volume of 22.4 dm3. (1 dm3 = 1 L)According to previous law, 1 mol - 22.4 dm3 x - 5.68 dm3-------------------------- x = 0.2536 mol1 mol of NO weights 30 g (Ar for N is 14, and Ar for O is 16. Mr(NO) = Ar(N) + Ar(O)), so1 mol - 30 g0.2536 mol - x------------------------------x = 7.6080 gSo the mass of 5.68 L of NO at STP is 7.6080 grams.
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
H3O= 0.9 mol/dm3 OH=1.2 mol/dm3
No.
ab
1 liter is 1 dm3
right i dont know this 4 sure but because u want a 0.1 mol/dm3 and u only need 100cm3 u will need 0.01mols of copper sulfate to dilute in 100cm3. soo now u have a solution that is 0.01mols per 100cm3 or 0.1 mols per 1000cm3 (dm3)
98 g per dm3 would be 1 molar - so 49 g would be 0.5 molar , so 4.9 g is 0.05 mol per dm3 or 0.05 mol.dm-3
dm3=1m2
100cm = 10 dm = 1m 1000 dm3 = 1 m3 ---> 786 dm3 = 0.786 m3
At STP 1 mole of every gas has the volume of 22.4 dm3. (1 dm3 = 1 L)According to previous law, 1 mol - 22.4 dm3 x - 5.68 dm3-------------------------- x = 0.2536 mol1 mol of NO weights 30 g (Ar for N is 14, and Ar for O is 16. Mr(NO) = Ar(N) + Ar(O)), so1 mol - 30 g0.2536 mol - x------------------------------x = 7.6080 gSo the mass of 5.68 L of NO at STP is 7.6080 grams.
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
kilolitres are 1000 litres, 1 dm3 = 1 litre so multiply by 1000
0.6 mol/dm3