The molar mass of anhydrous MgSO4 is 120,366 g.
2 NaOH + MgSO4-----Mg(OH)2+Na2SO4 1 MOLE OF MgSO4 produce one mole of Mg(OH)2 SO 1 MWT OF MgSO4 PRODUCE1MWT OF Mg(OH)2 HENCE 120 GM MgSO4 PRODUCE 58 gm OF Mg(OH)2 and so on if x gm of MgSO4 PRODUCE y gm OF Mg(OH)2 x=(120*y)/58 ---- ----
Only MgSO4 is a correct formula.
The molecular weight of MgSO4 is 120.3676 g/mol.
No; 1 mole of molcular oxygen (O2) is 31,998 g and 1 mole of sulfur (S) is 32,06 g.
No Magnesium Sulfate (MgSO4) has a mass of 121, and Sodium Chloride (NaCl) has a mass of 59.
gh
Mass of 1mole of Argon is 39.95g. So, The mass of 7 moles of Argon is 39.95x7= 279.65g.
The molar mass of anhydrous MgSO4 is 120,366 g.
The relative atomic mass of an element is the mass in grams of 1mole of the substance. The relative atomic mass of copper is 63.5 2x63.5 = 127g
54.94g Manganese = 1mole = 6.023 x 1023 atoms
48.8 g MgSO4 & 51.2 g H2O Convert the mass into moles by dividing molar mass for each. Then obtain a ratio of moles of water over moles of Magnesium Sulfate, and you would get 7. MgSO4 . 7H2O would read as Magnesium Sulfate Heptahydrate.
2 NaOH + MgSO4-----Mg(OH)2+Na2SO4 1 MOLE OF MgSO4 produce one mole of Mg(OH)2 SO 1 MWT OF MgSO4 PRODUCE1MWT OF Mg(OH)2 HENCE 120 GM MgSO4 PRODUCE 58 gm OF Mg(OH)2 and so on if x gm of MgSO4 PRODUCE y gm OF Mg(OH)2 x=(120*y)/58 ---- ----
MgSO4
MgSO4 is not binary.
First we calculate the formula mass of the compound magnesium sulfate.Formula mass of MgSO4 = 24.3 + 32.1 + 4(16.0) = 120.4 Amount of MgSO4 in a 480g pure sample = 480/120.4 = 3.99mol There is approximately 4 moles of the compound present in a 480g sample.
The formula for caffeine is C8H10N4O2 and its molar mass is 194.19g/mole. 1mol caffeine = 194.19g 0.00100mole caffeine X (194.19g/1mole) = 0.19419g caffeine