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.
Answer:
B protons determine indentity and valence electrons determine chemical properties.
:)
Explanation:
Nitrate
Hope this helps :))
Answer:
The final temperature is 348.024°C.
Explanation:
Given data:
Specific heat of copper = 0.385 j/g.°C
Energy absorbed = 7.67 Kj (7.67×1000 = 7670 j)
Mass of copper = 62.0 g
Initial temperature T1 = 26.7°C
Final temperature T2 = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
Q = m.c. ΔT
7670 J = 62.0 g × 0.385 j/g °C ×( T2- 26.7 °C
)
7670 J = 23.87 j.°C ×( T2- 26.7 °C
)
7670 J / 23.87 j/°C = T2- 26.7 °C
T2- 26.7 °C = 321.324°C
T2 = 321.324°C + 26.7 °C
T2 = 348.024°C
The final temperature is 348.024°C.