Answer:

Explanation:
Assume you are using 1 L of water.
Then you are washing 4 L of salty oil.
1. Calculate the mass of the salty oil
Assume the oil has a density of 0.86 g/mL.

2. Calculate the mass of salt in the salty oil

3. Calculate the mass of salt in the spent water

4. Mass of salt remaining in washed oil
Mass = 172 g - 150 g = 22 g
5. Concentration of salt in washed oil

Answer:
Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)
Explanation:
Molecular Equation:
Zn°(s) + Fe(NO₃)₂(aq) => Zn(NO₃)₂(aq) + Fe°(s)
Ionic Equation:
Zn°(s) + Fe⁺²(aq) + 2NO₃⁻(aq) => Zn⁺²(aq) + 2NO₃⁻(aq) + Fe°(s)
Net Ionic Equation: => Drop NO₃⁻ as spectator ion
Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)
Answer:
The equation for the reaction of one sodium bicarbonate ( NaHCO3 ) molecule with one citric acid (C6H8O7) molecule is the following:
Sodium Bicarbonate + Citric Acid ⇒ Water + Carbon Dioxide + Sodium Citrate
NaHCO3 + C6H8O7 ⇒ 3 CO2 + 3 H2O + Na3C6H5O7
Explanation:
The reaction is in balance, that is, the whole H2CO3 is not finished, but a little bit of this acid is left in the solution. Therefore, when sodium bicarbonate is added to the solution with citric acid, sodium citrate salt (C6H5O7Na3) and carbonic acid (H2CO3) are formed, which is rapidly broken down into water (H2O) and carbonic oxide (CO2).
C6H8O7 + NaHCO3 ⇒ C6H5O7Na3 + 3 H2CO3
C6H5O7Na3 + 3 H2CO3 ⇔ C6H5O7Na3 + 3 H2O + 3 CO2
Answer : The mass of 7.0 m chain is, 15.12 kg
Explanation :
As we are given that,
The weight of the chain per unit length = 2.16 kg/m
Now we have to determine the mass of chain for 7.0 m length.
As, the mass of 1 m length of chain = 2.16 kg
So, the mass of 7.0 m length of chain = 
= 15.12 kg
Therefore, the mass of 7.0 m chain is, 15.12 kg
Answer : The molar concentration of ethanol in the undiluted cognac is 8.44 M
Explanation :
Using neutralization law,

where,
= molar concentration of undiluted cognac = ?
= molar concentration of diluted cognac = 0.0844 M
= volume of undiluted cognac = 5.00 mL = 0.005 L
= volume of diluted cognac = 0.500 L
Now put all the given values in the above law, we get molar concentration of ethanol in the undiluted cognac.


Therefore, the molar concentration of ethanol in the undiluted cognac is 8.44 M