Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
Lithium 6 would have 6 valence electrons in the outer orbital, while lithium 7 would have 7 in the outer orbital.
Answer:
- smoke
- marshmallow
- paint
- Body spray
Explaination:
Firstly, what are colloids?
Colloids are a mixture of two substances:
- one of those substances are insoluble and inseparable
There are 4 types of colloids:
- Sol- Solid in a liquid
- Emulsion- Liquid in a liquid
- Foam- Gas in solid/liquid
- Aerosol- Solid/liquid in gas
So on to the answers:
- Smoke consists of solid and gas and since the solid is being carried by the gas, it is a solid in a gas (aerosol)
- Marshmellow consists of tiny air bubbles and sugar(solid), the air bubbles is inside the solid making it a gas in a solid (foam)
- Body spray turns gas into liquid after spraying it through the can, making it a liquid in a gas (aerosol)
- Paint dries up to become solid which makes it a solid is a liquid (sol)
Hope y'all understand
Answer:
92.49 %
Explanation:
We first calculate the number of moles n of AgBr in 0.7127 g
n = m/M where M = molar mass of AgBr = 187.77 g/mol and m = mass of AgBr formed = 0.7127 g
n = m/M = 0.7127g/187.77 g/mol = 0.0038 mol
Since 1 mol of Bromide ion Br⁻ forms 1 mol AgBr, number of moles of Br⁻ formed = 0.0038 mol and
From n = m/M
m = nM . Where m = mass of Bromide ion precipitate and M = Molar mass of Bromine = 79.904 g/mol
m = 0.0038 mol × 79.904 g/mol = 0.3036 g
% Br in compound = m₁/m₂ × 100%
m₁ = mass of Br in compound = m = 0.3036 g (Since the same amount of Br in the compound is the same amount in the precipitate.)
m₂ = mass of compound = 0.3283 g
% Br in compound = m₁/m₂ × 100% = 0.3036/0.3283 × 100% = 0.9249 × 100% = 92.49 %
