Answer:
We need 7.5 mL of the 1M stock of NaCl
Explanation:
Data given:
Stock = 1M this means 1 mol/ L
A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL
Step 2: Calculate the volume of stock we need
The moles of solute will be constant
and n = M*V
M1*V1 = M2*V2
⇒ with M1 = the initial molair concentration = 1M
⇒ with V1 = the volume we need of the stock
⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L
⇒ with M2 = the concentration of the new solution = 0.15 M
1*V1 = 0.15*(50)
V1 = 7.5 mL
Since 0.0075 L of 1M solution contains 0.0075 moles
50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M
We need 7.5 mL of the 1M stock of NaCl
Answer:
1) pure water
2) 0.75 m CaCl2
3) 1.0 m NaCl
4) 0.5 m KBr
5) 1.5 m glucose (C6H12O6)
Explanation:
Boiling point elevation is a colligative property. Coligative properties are properties that depend on the amount of solute present in the system. The boiling point of solvents increase due to the presence of solutes.
The boiling point elevation depends on the number of particles the solute forms in solution and the molality of the solute. The more the number of particles formed by the solute and the greater the molality of the solute, the greater the magnitude of boiling point elevation.
The order of decreasing hoping point elevation is;
1) 0.75 m CaCl2
2) 1.0 m NaCl
3) 0.5 m KBr
4) 1.5 m glucose (C6H12O6)
Answer:
2Mg + O₂ → 2MgO
Explanation:
Chemical equation:
Mg + O₂ → MgO
Balanced chemical equation:
2Mg + O₂ → 2MgO
The balanced equation s given above and it completely follow the law of conservation of mass.
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Steps to balanced the equation:
Step 1:
Mg + O₂ → MgO
Mg = 1 Mg = 1
O = 2 O = 1
Step 2:
2Mg + O₂ → MgO
Mg = 2 Mg = 1
O = 2 O = 1
Step 3:
2Mg + O₂ → 2MgO
Mg = 2 Mg = 2
O = 2 O = 2