M = n/V
.5M = n/.100 L
n = .1 L * .5M
n= .05 mols of MgCl2
mass of MgCl2 = .05 mols of MgCl2 * 95.211 grams/ 1 mol of MgCl2
mass of MgCl2 = 4.76 grams
4.76 grams of MgCl2 is needed to make 100 ml of a solution that is .500M, in chloride ion. Bolded = confused
Answer:
The mixture of ethanol and water is a liquid-liquid solution. The solvent is Water.
Explanation:
Answer:
25.89 × 10²³ molecules
Explanation:
Given data:
Mass of CoCl₂ = 560 g
Number of molecules present = ?
Solution:
Number of moles of CoCl₂:
Number of moles = mass/molar mass
Number of moles = 560 g/ 129.84 g/mol
Number of moles = 4.3 mol
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
4.3 mol × 6.022 × 10²³ molecules /1 mol
25.89 × 10²³ molecules
Mg + 1/2 O2 → MgO
1 mol = 24 g of Mg
X mol = 12 g of Mg
x = 0.5 moles of Mg
Mg :MgO = 1:1 (coefficient from equations using mole ratio)
So
0.5 moles of MgO
1 mol MgO = (24+16) g = 40 g
0.5 moles of MgO = 0.5 × 40
= 20 g of MgO produced
