Answer:
Density = 4.191 gm/L
Explanation:
Given:
Molar mass = 93.89 g/mol
Volume(Missing) = 22.4 L (Approx)
Find:
Density at STP
Computation:
Density = Mass/Volume
Density = 93.89 / 22.4
Density = 4.191 gm/L
The value of X is 10 hence the formula of unknown hydrate sodium sulfate is NaSO4.10 H20
calculation
step 1:find the moles of NaSO4 and the moles of H2O
moles= mass/molar mass
moles of Na2SO4=1.42÷142=0.01 moles
moles of H20= mass of H2O/molar mass of H2O
mass of H2O= 3.22-1.42=1.8g
mole of H2O is therefore 1.8÷18=0.1 moles
step 2: find the mole ratio by dividing each mole by smallest number of mole (0.01)
that is Na2So4= 0.01/0.01 =1
H2O= 0.1/0.01=10
Answer:
Option c → Tert-butanol
Explanation:
To solve this, you have to apply the concept of colligative property. In this case, freezing point depression.
The formula is:
ΔT = Kf . m . i
When we add particles of a certain solute, temperature of freezing of a solution will be lower thant the pure solvent.
i = Van't Hoff factor (ions particles that are dissolved in the solution)
At this case, the solute is nonvolatile, so i values 1.
ΔT = Difference between fussion T° of pure solvent - fussion T° of solution.
T° fussion paradichlorobenzene = 56 °C
T° fussion water = 0°
T° fussion tert-butanol = 25°
Water has the lowest fussion temperature and the paradichlorobenzene has the highest Kf. But the the terbutanol, has the highest Kf so this solvent will have the largest change in freezing point, when all the molalities are the same.
Answer:
1219.5 kj/mol
Explanation:
To reach this result, you must use the formula:
ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)
ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).
The BE values are:
BE C = C: 839 kj / mol
BE C-H: 413 Kj / mol
BE O = O: 495 kj / mol
BE C = O = 799 Kj / mol
BE O-H = 463 kj / mol
Now you must replace the values in the above equation, the result of which will be:
ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol
The roots could no longer access depleted groundwater.
The topsoil in the area eroded.