first work out how many moles of Mg(OH)2 you have.
n = m /M
= 3.33 / 58 = 0.0574 mol
Then from this work out how many mol of O you have.
there are two oxygens in the formula so you have 0.0574 x 2 mol of O = 0.1148 mol
1 mol = 6.023x 1023 particles.
So 6.023 x 10 23 /1 x 0.1148 = 6.92 x 10 22
i dont know help :C
Magnesium chloride: 81,95 g are obtained.
24.31 grams of magnesium is one mole of magnesium, so that is 6.022 X 1023 atoms of magnesium.
2.04 g
1
340 grams
how much sodium hydroxide in grams must be added to seawater to precipitate 86.9mg of magnesium present?
i dont know help :C
2
Magnesium chloride: 81,95 g are obtained.
24.31 grams of magnesium is one mole of magnesium, so that is 6.022 X 1023 atoms of magnesium.
A lot
3,45 grams of H2O contain 1,154.10e23 oxygen atoms.
Mass of Oxygen is approximately 16 grams. (15.9994 grams) per mole. Mass of Magnesium is approximately 24.305 grams per mole.
3.120 grams oxygen gas (1 mole O2/32 grams)(6.022 X 1023/1 mole O2) = 5.871 X 1022 atoms of oxygen gas ------------------------------------------------
Let us find Magnesium atoms first. 15 grams magnesium (1 mole Mg/24.31 grams)(6.022 X 10^23/1 mole Mg) = 3.72 X 10^23 atoms magnesium Now take this an drive to grams sodium 3.72 X 10^23 atoms (1 mole Na/6.022 X 10^23)(22.99 grams/1 mole Na) = 14 grams of sodium ----------------------------
2.04 g