Answer:
1) 90.0 mL
2) 11.25 M
3) 0.477 M
4) 144 mL
Explanation:
The main formula that will be used for all these calculations is:
C₁V₁ = C₂V₂
C stands for concentration and V stands for volume and the subscripts 1 and 2 indicate an initial concentration or volume and a final concentration or volume.
For each problem, it's best to start by figuring out what you have and what you need to find. Figure out if you're looking for an initial value or a final value.
1) We need to find the initial volume. So, take what values you have and plug them in and then solve for whatever variable:
5.00 M · V₁ = 500.0mL · 0.900 M - divide by 5.00
C₁ = 90.0 mL
2) This time we're finding the initial concentration:
20.0mL · C₁ = 150.0mL · 1.50 M - divide by 20.0mL
C₂ = 11.25 M
3) Now we're finding the final concentration:
12.00mL · 3.50 M = 88.0mL · C₂ - divide by 88.0mL
C₂ = 0.477 M
4) Finally, we're looking for the final volume:
9.0mL · 8.0 M = 0.50 M · V₂ - divide by 0.50 M
V₂ = 144mL
<u>Answer:</u> The volume of concentrated solution required is 9.95 mL
<u>Explanation:</u>
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:
pH = 0.70
Putting values in above equation, we get:
![0.70=-\log[H^+]](https://tex.z-dn.net/?f=0.70%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-0.70}=0.199M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-0.70%7D%3D0.199M)
1 mole of nitric acid produces 1 mole of hydrogen ions and 1 mole of nitrate ions.
Molarity of nitric acid = 0.199 M
To calculate the volume of the concentrated solution, we use the equation:
where,
are the molarity and volume of the concentrated nitric acid solution
are the molarity and volume of diluted nitric acid solution
We are given:
Putting values in above equation, we get:
Hence, the volume of concentrated solution required is 9.95 mL
It should be organized in people
Answer:
87.3 calories of heat is required.
Explanation:
Heat = mcΔT
m= mass, c = specific heat of silver, T = temperature
H= 57.8 g * 0.057 cal/g°C * ( 43.5 - 17 °C)
H = 57.8 * 0.057 * 26.5
H = 87.3069 cal.
The heat required to raise the temperature of 57.8 g of silver from 17 °C to 43.5 °C is 87.3 calories.
