3.5 M has 3.5 moles per litre
so we have one litre, so we need 3.5 moles
moles = mass/molarmass
3.5 * 23 = 80.5
An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.
Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]
moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles
moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles
pKa of phenol = 9.98
We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation
pH = pKa + log [salt] / [acid]
volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula
pH = 9.98 + log [12.93 / 42.55] = 9.46
The reaction is:
X + Y → W + Z
Chemical energy of reactant X = 199.3 J = Ux
Chemical energy of reactant Y = 272.3 J = Uy
Chemical energy of Product W = 41.9 J = Uw
Chemical energy of Product Z = ? = Uz
Where reaction loses energy = 111.6 J = ΔU
By using the equation:
(Ux + Uy) – (Uw + Uz) = ΔU
Ux + Uy – Uw – Uz = ΔU
Uz = Ux + Uy – Uw –ΔU
Uz = 199.3 + 272.3 – 41.9 – 111.6
Uz = 318.1 J
Product Z must contain 318.1 J chemical energy.
Answer:We are already given with the mass of the Xe and it is 5.08 g. We can calculate for the mass of the fluorine in the compound by subtracting the mass of xenon from the mass of the compound.
mass of Xenon (Xe) = 5.08 g
mass of Fluorine (F) = 9.49 g - 5.08 g = 4.41 g
Determine the number of moles of each of the element in the compound.
moles of Xenon (Xe) = (5.08 g)(1 mol Xe / 131.29 g of Xe) = 0.0387 mols of Xe
moles of Fluorine (F) = (4.41 g)(1 mol F/ 19 g of F) = 0.232 mols of F
The empirical formula is therefore,
Xe(0.0387)F(0.232)
Dividing the numerical coefficient by the lesser number.
XeF₆
Explanation:
Answer:
1.67×10^25 molecules
Explanation:
No of molecules = no of moles × Avogadros number
No of moles= mass in gram / molar mass
No of moles of water in given sample = 500.3/18
= 27.79 moles
No of molecules = 27.79× 6.02×10^ 23
= 167.32×10^23 or 1.67×10^25
