The concentrations : 0.15 M
pH=11.21
<h3>Further explanation</h3>
The ionization of ammonia in water :
NH₃+H₂O⇒NH₄OH
NH₃+H₂O⇒NH₄⁺ + OH⁻
The concentrations of all species present in the solution = 0.15 M
Kb=1.8 x 10⁻⁵
M=0.15
![\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7BKb.M%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.15%7D%5C%5C%5C%5C%28OH%5E-%5D%3D%5Csqrt%7B2.7%5Ctimes%2010%5E%7B-6%7D%7D%3D1.64%5Ctimes%2010%5E%7B-3%7D)
![\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21](https://tex.z-dn.net/?f=%5Ctt%20pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D3-log~1.64%3D2.79%5C%5C%5C%5CpH%3D14-2.79%3D11.21)
Hey there!:
Molar mass of Mg(OH)2 = 58.33 g/mol
number of moles Mg(OH)2 :
moles of Mg(OH)2 = 30.6 / 58.33 => 0.5246 moles
Molar mass of H3PO4 = 97.99 g/mol
number of moles H3PO4:
moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles
Balanced chemical equation is:
3 Mg(OH)2 + 2 H3PO4 ---> Mg3(PO4)2 + 6 H2O
3 mol of Mg(OH)2 reacts with 2 mol of H3PO4 ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !
Now , we will use Mg(OH)2 in further calculation .
Molar mass of Mg3(PO4)2 = 262.87 g/mol
According to balanced equation :
mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2
= (1/3)*0.5246
= 0.1749 moles of Mg3(PO4)2
use :
mass of Mg3(PO4)2 = number of mol * molar mass
= 0.1749 * 262.87
= 46 g of Mg3(PO4)2
Therefore:
% yield = actual mass * 100 / theoretical mass
% = 34.7 * 100 / 46
% = 3470 / 46
= 75.5%
Hope that helps!
Answer:
The groupings of elements in the periodic table reveal various trends, present in groups such as the alkali metals and noble gases, or the lanthanides and actinides. The periodic table can be used to find information on elements or predict the properties of as-of-yet undiscovered elements.
Explanation:
Answer:
a) Ka= 7.1 × 10⁻⁴; This is a weak acid because the acid is not completely dissociated in solution.
Explanation:
Step 1: Write the dissociation reaction for nitrous acid
HNO₂(aq) ⇄ H⁺(aq) and NO₂⁻(aq)
Step 2: Calculate the acid dissociation constant
Ka = [H⁺] × [NO₂⁻] / [HNO₂]
Ka = 0.022 × 0.022 / 0.68
Ka = 7.1 × 10⁻⁴
Step 3: Determine the strength of the acid
Since Ka is very small, nitrous acid is a weak acid, not completely dissociated in solution.
We are given with the mass of pure iron that reacts with oxygen to form an oxide which has a given mass as well. the mass of oxygen reacted is 8.15-6.25 g or 1.9 grams. THen we convert the mass of the reactants to moles. Iron is equal to 0.1119 moles and oxygen is equal to 0.1188. We divide each number to the less amount. Hence iron is 1 and oxygen is approx 1. The empirical formula hence is FeO or ferrous oxide or Iron (II) oxide.
