Answer:
3rd or 4th answer is correct
Answer: -
15.55 M
35.325 molal
Explanation: -
Let the volume of the solution be 1000 mL.
Density of nitric acid = 1.42 g/ mL
Total Mass of nitric acid Solution = Volume of nitric acid x Density of nitric acid
= 1000 mL x 1.42 g/ mL
= 1420 g.
Percentage of HNO₃ = 69%
Amount of HNO₃ = 
= 979.8 g
Molar mass of HNO₃ = 1 x 1 + 14 x 1 + 16 x 3 = 63 g /mol
Number of moles of HNO₃ = 
= 15.55 mol
Molarity is defined as number of moles per 1000 mL
We had taken 1000 mL as volume and found it to contain 15.55 moles.
Molarity of HNO₃ = 15.55 M
Mass of water = Total mass of nitric acid solution - mass of nitric acid
= 1420 - 979.8
= 440.2 g
So we see that 440.2 g of water contains 15.55 moles of HNO₃
Molality is defined as number of moles of HNO₃ present per 1000 g of water.
Molality of HNO₃ = 
= 35.325 molal
Complete Question
A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane
The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?
Answer:
The mass of caffeine extracted is 
Explanation:
From the question above we are told that
The K value is 
The mass of the caffeine is 
The volume of water is 
The volume of caffeine is 
The number of times the extraction was done is n = 3
Generally the mass of caffeine that will be extracted is
![P = m * [\frac{V}{K * v_c + V} ]^3](https://tex.z-dn.net/?f=P%20%3D%20%20m%20%20%2A%20%20%5B%5Cfrac%7BV%7D%7BK%20%2A%20%20v_c%20%2B%20V%7D%20%5D%5E3)
substituting values
![P = 40 * [\frac{2}{4.6 * 2 + 2} ]^3](https://tex.z-dn.net/?f=P%20%3D%20%2040%20%20%20%2A%20%20%5B%5Cfrac%7B2%7D%7B4.6%20%2A%20%202%20%2B%202%7D%20%5D%5E3)

Answer:
20 molecules of oxygen gas remains after the reaction.
Explanation:

Molecules of ethyne = 52
Molecules of oxygen gas = 150
According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.
Then 52 molecules of ethyne will react with:
of oxygen gas.
As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.
Remaining molecules of recessive reagent = 150 - 130 = 20
20 molecules of oxygen gas remains after the reaction.