Answer:
60 cm³ of water
Explanation:
We'll begin by calculating the volume of the diluted solution. This can be obtained as follow:
Concentration of stock solution (C₁) = 17 M
Volume of stock solution (V₁) = 25 cm³
Concentration of diluted solution (C₂) = 5 M
Volume of diluted solution (V₂) =?
C₁V₁ = C₂V₂
17 × 25 = 5 × V₂
425 = 5 × V₂
Divide both side by 5
V₂ = 425 / 5
V₂ = 85 cm³
Thus, the volume of the diluted solution is 85 cm³
Finally, we shall determine the volume of water needed to dilute the solution. This can be obtained as follow:
Volume of stock solution (V₁) = 25 cm³
Volume of diluted solution (V₂) = 85 cm³
Volume of water =?
Volume of water = V₂ – V₁
Volume of water = 85 – 25
Volume of water = 60 cm³
Therefore, 60 cm³ of water is needed to dilute the solution.
Answer:
300 mL
Explanation:
the unit formula of calcium phosphate is Ca3(PO4)2
molar mass of Ca3(PO4)2 = (3×40 + 2×31 + 8×16) g/mol = 310 g/mol
n = m/M = 35 g/(310 g/mol)
c = n/V
V = n/c = [35 g/(310 g/mol)]/0.375 mol/L
V = 0.30 L = 300 mL
Answer:
Spanish
Explanation:
help is urgent the log of wood and the same
The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.
To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.
From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.
1) From pV = nRT, n = pV / RT
Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K
n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol * 308.15K] =0.00472 mol
mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042
=> MM = mass/n = 0.042 / 0.00472 = 8.90 g/mol
Now from a periodic table or a table you get that the molar mass of He is 4g/mol
So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.
