Answer:
Explanation:
You don't give the reaction, but we can get by just by balancing atoms of Na.
We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
M_r: 142.04
2NaOH + … ⟶ Na₂SO₄ + …
n/mol: 0.75
1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.
Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH
= 0.375 mol Na₂SO₄
2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.
Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄
The reaction produces of Na₂SO₄.
Answer: Reactivity is not a physical property, it is a chemical property.
Explanation:
Physical property is a value of matter that can be observed and measured without any change in composition of the matter. Examples of physical properties are color, combustion,melting point,boiling point,solubility, texture, appearance, polarity, odor,viscosity, density and many more.
Chemical properties are characteristics that are obvious or seen when substances undergo chemical reaction. Examples are reactivity,toxicity, acidity, flammability, and heat combustion.
Answer:
atoms Ti = 2.193 E23 atoms
Explanation:
- 1 mol ≡ 6.022 E23 atoms
- mm Ti = 47.867 g/mol
⇒ atoms Ti = (20.9 g Ti)×(mol/47.867 g)×(6.022 E23 atoms/mol)
⇒ atoms Ti = 2.193 E23 atoms
Answer:
130.4 grams of sucrose, would be needed to dissolve in 500 g of water.
Explanation:
Colligative property of boiling point elevation:
ΔT = Kb . m . i
In this case, i = 1 (sucrose is non electrolytic)
ΔT = Kb . m
0.39°C = 0.512°C/m . m
0.39°C /0.512 m/°C = m
0.762 m (molality means that this moles, are in 1kg of solvent)
If in 1kg of solvent, we have 0.712 moles of sucrose, in 500 g, which is the half, we should have, the hallf of moles, 0.381 moles
Molar mass sucrose = 342.30 g/m
Molar mass . moles = mass
342.30 g/m . 0.381 m = 130.4 g
