Answer:
V₁ = 374.71 mL
Explanation:
Given data:
Initial volume of gas= ?
Initial temperature = 22°C
Final temperature = 86°C
Final volume = 456 mL
Solution:
Initial temperature = 22°C (22+273 = 295 k)
Final temperature = 86°C (86+273 = 359 k)
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₁ = V₂T₁ /T₂
V₁ = 456 mL × 295 K / 359 k
V₁ = 134520 mL.K / 359 k
V₁ = 374.71 mL
2.25 I believe
Hope this helped!
STSN
Answer:
The correct answer is option c.
Explanation:
Formula used to determine an average atomic mass :
Mass of isotope Sb-121 = 120.904 amu
Fractional abundance of Sb-121 = 57.21% = 0.5721
Mass of isotope Sb-123 = 122.904 amu
Fractional abundance of Sb-123 = 42.79% = 0.4279
Average atomic mass of Sb:
Answer:
The volume of water to be added is 0.175 liters of water
Explanation:
The given concentration of the nitric acid = 55% (M/M)
The mass of the nitric acid solution = 100 gm
The concentration solution is to diluted to = 20% (M/M)
The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution
Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get
Let "x" represent the volume of the resulting solution, we have;
20% of x = 55 g of nitric acid
∴ 20/100 × x = 55 g
x = 55 g × 100/20 = 275 g
The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid
The mass of extra water to be added = 275 g - 100 g = 175 g
Volume = Mass/Density
The density of water ≈ 1 g/ml
∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l
The volume of water to be added = 0.175 liters of water.
