To determine the volume of both concentration of vinegar, we need to set up two equations since we have two unknowns.
For the first equation, we do a mass balance:
mass of 100% vinegar + mass of 13% vinegar = mass of 42% vinegar
Assuming they have the same densities, then we can write this equation in terms of volume.
V(100%) + V(13%) = V(42%)
we let x = V(100%)
y = V(13%)
x + y = 150
For the second equation, we do a component balance:
1.00x + .13y = 150(.42)
x + .13y = 63
The two equations are
x + y = 150
x + .13y = 63
Solving for x and y,
x = 50
y = 100
Therefore, you need to mix 50 mL of the 100% vinegar and 100 mL of the 13% vinegar.
Answer:
2,909 M
Explanation:
molair mass is of.ethylene is 26,04 g/mol
first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.
3000 g x 1.1 = 3300 mL = 3,3 L
Next you need to calculate the amount of moles:
250 g / 26,04 g/mol = 9,60 mol
Now you can calculate the molarity:
9,6/3.3 = 2,909 M
I don't know the answer for the second question. I'm sorry.
<u>Answer:</u> The initial pH of the HCl solution is 3
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
1 mole of HCl produces 1 mole of 
ions and 1 mole of 
ions
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
![[H^+]=0.001M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.001M)
Putting values in above equation, we get:
Hence, the initial pH of the HCl solution is 3
Answer: chemical
Explanation: whenever heat is applied, it indicates a chemical change
