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Answer:
11.6 grams of MgO can be produced with 4.63 grams of oxygen.
Explanation:
Mass of oxygen gas = 4.63 g
Moles of oxygen gas =
According to reaction, 1 mole of oxygen gas gives 2 moles of magnesium oxide, then 0.1447 moles of oxygen gas will giv e:
of magnesium oxide
Mass of 0.2894 moles of magnesium oxide:
= 0.29 mol × 40 g/mol = 11.6 g
11.6 grams of MgO can be produced with 4.63 grams of oxygen.
Answer:
The pH of saturated solution of the quinine is 10.05
Explanation:
Quinine (Q) is C20H24N2O2 has a molar mass of 324.4 g/mol
Q can behave as a weak base. Kb and pKb can be calculated for weak bases
pKb1 is to be considered when solving the question.
pKb1 = 5.1
Step 1 : Calculate the Kb of Quinine
pKb1 = - log [kb]
5.1 = - log [kb]
take Antilog of both side
[kb] = 7.94 x 10∧-6
Step 2: Calculate the concentration of saturated solution of Q in mol/dm3
From the question, 1900 ml of solution contains 1 g of Q
Therefore, 1000 ml of solution will contain........... x g of Q
x = 1000 /1900
x = 0.526 g in 1 dm3
In calculating concentration in mol/dm3,
Concentration in mol/dm3 = concentration in g /dm3 divided by molar mass
Molar mass of Q = 324.4
Concentration in mol/dm = 0.526 /324.4
= 0.0016 mol/dm3
Step 3: Calculating the Concentration of OH-
At Equilibrium, Kb = x² / 0.0016
7.94 x 10∧-6 = x² / 0.0016
x = √ 0.0016 × 7.94 x 10∧-6
x = 1. 127 × 10∧-4 mol/dm3
The concentration of OH- = 1. 127 × 10∧-4 mol/dm
Step 4: Calculating the pH of Quinine
Recall, pOH = - log [OH-]
pOH = - log [1. 127 × 10∧-4]
pOH = 3.948
Also recall that pH + pOH = 14
pH = 14 - 3.948
pH = 10.05
