The empirical formula : MnO₂.
<h3>Further explanation</h3>Given
632mg of manganese(Mn) = 0.632 g
368mg of oxygen(O) = 0.368 g
M Mn = 55
M O = 16
Required
The empirical formula
Solution
You didn't include the pictures, but the steps for finding the empirical formula are generally the same
Mn : 0.632 : 55 = 0.0115
O : 0.368 : 16 =0.023
Mn : O =
The empirical formula : MnO₂
Answer:
The volume of the sample of the gas is found to be 12.90 L.
Explanation:
Given pressure of the gas = P = 1.10 atm
Number of moles of gas = n = 0.6000 mole
Temperature = T = 288.15 K
Assuming the volume of the gas to be V liters
The ideal gas equation is shown below
Volume occupied by gas = 12.90 L
Answer:
pH = 9,57
[M²⁻] = 7,948x10⁻²M
[HM⁻] = 4x10⁻⁵M
[H₂M] = 0M
Explanation:
The moles of maleic acid presents in the solution are:
0,923g×=7,952x10⁻³moles of H₂M
60,0mL of 0,265M KOH are:
0,0600L×=0,0159 moles of KOH
The reactions of maleic acid (H₂M) and then with HM⁻ are:
H₂M + KOH → HM⁻ + H₂O + K⁺ (1)
HM⁻ + KOH → M²⁻ + H₂O + K⁺ (2)
For a complete transformation of H₂M in HM⁻ there are necessaries 7,952x10⁻³moles of KOH. As the moles of KOH are 0,0159 moles, the restant moles are:
0,0159 - 7,952x10⁻³ = <em>7,948x10⁻³ moles of KOH</em>
By (2), the moles produced of M²⁻ are the same as moles of KOH, <em>7,948x10⁻³ moles, </em>and moles of HM⁻ are:
7,952x10⁻³ -<em> </em>7,948x10⁻³ <em> = 4x10⁻⁶ moles of HM⁻</em>
Using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [M²⁻] /[HM⁻]
pH = 6,27 + log₁₀ <em>7,948x10⁻³ / 4x10⁻⁶</em>
pH = 9,57
The moles of M²⁻ are 7,948x10⁻³ and volume of the solution is 0,1000L,
[M²⁻] = 7,948x10⁻²M
Moles of HM⁻ are 4x10⁻⁶:
[HM⁻] = 4x10⁻⁵M
And there is not H₂M:
[H₂M] = 0M
I hope it helps!