Answer:
Explanation:
Since water has a chemical formula of H2O , there will be 2 moles of hydrogen in every mole of water. In one mole of water, there will exist approximately 6.02⋅1023 water molecules.
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
Given mass of KNO₃=346g
Molar mass of KNO₃=(39.098)+(14)+(15.99*3)=101.068gmol⁻¹
Volume of Solution=750ml=0.75dm³
Molarity=(mass of solute/molar mass of solute)*(1/volume of sol. in dm³)
=(346/101.068)*(1/0.75)
=4.56 mol dm⁻³
Find moles of MgSO4.7H2O
molar mass = 246
so moles = 32 / 246 = 0.13 moles.
When heated, all 7 H2O from 1 molecule will be gone.
total moles of H2O present = 7 x 0.13 = 0.91
mass of those H2O = 0.91 x 18 = 16.38g
so mass of anyhydrous MgSO4 remain = 32 - 16.38 = 15.62 g
Answer:
The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>
Explanation:
Given: The base dissociation constant: 
= 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M
Also, water dissociation constant: 
= 1 × 10⁻¹⁴
<em><u>The acid dissociation constant </u></em>(
)<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>
<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>
Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+
Initial: 0.1 M x x
Change: -x +x +x
Equilibrium: 0.1 - x x x
<u>The acid dissociation constant: </u>![K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5Cleft%20%5BB%20%5Cright%20%5D%20%5Cleft%20%5BH_%7B3%7DO%5E%7B%2B%7D%5Cright%20%5D%7D%7B%5Cleft%20%5BBH%5E%7B%2B%7D%20%5Cright%20%5D%7D%20%3D%20%5Cfrac%7B%28x%29%28x%29%7D%7B%280.1%20-%20x%29%7D%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%7D%7B0.1%20-%20x%7D)
<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>
Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44
<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>
