Answer:
1. 8.7moles of H2
2. 2.25moles of O2
Explanation:
1. 2NH3 —> N2 + 3H2
From the equation,
2moles of NH3 produce 3 moles of H2.
Therefore, 5.8moles of NH3 will produce Xmol of H2 i.e
Xmol of H2 = (5.8x3)/2 = 8.7moles
2. C3H8 + 5O2 —> 3CO2 + 4H2O
From the equation,
5moles of O2 produced 4moles of H2O.
Therefore, Xmol of O2 will produce 1.8mol of H2O i.e
Xmol of O2 = (5x1.8)/4 = 2.25moles
A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the
solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.
1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.
Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.
It is calculated in mg/ml.
The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.
Formula for calculating concentration in mg/ml is
Conc. (mg/ml) = M(eq) /ml × Molecular weight / Valency
Given
M(eq) NaCl/ ml = 23.5
Molecular weight pf NaCl = 58.5 g/mol
Valency = 1
Putting the values into the formula
Conc. (mg/ml) = 23.5 ×58.5/1
= 1374.75 mg/ml
Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.
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