Answer:
<u><em>PLEASE SHOW ME THE ELEMENTS OR I WOULD ENLIST ALL THE ELEMENTS.</em></u>
Explanation:
Group 6A (or VIA) of the periodic table are the chalcogens: the nonmetals oxygen (O), sulfur (S), and selenium (Se), the metalloid tellurium (Te), and the metal polonium (Po)
Answer is: <span>the percent ionization is 0,19%.
</span>Chemical reaction: HA(aq) ⇄ H⁺(aq) + A⁻(aq).
Ka(HA) = 3,6·10⁻⁷.
c(HA) = 0,1 M.
[H⁺] = [A⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [A⁻] / [HA].
0,00000036 = x² / 0,1 M - x.
Solve quadratic equation: x = 0,00019 M.
α = 0,00019 M ÷ 0,1 M · 100% = 0,19%.
Answer:
0.3267 M
Explanation:
To solve this problem, first we calculate how many moles of Mn(ClO₄)₂ are contained in 23.640 g of Mn(ClO₄)₂·6H₂O.
Keep in mind that the crystals of Mn(ClO₄)₂ are hydrated, and <em>we need to consider those six water molecules when calculating the molar mass of the crystals</em>.
Molar mass of Mn(ClO₄)₂·6H₂O = 54.94 + (35.45+16*4)*2 + 6*18 = 361.84 g/mol
Now we <u>proceed to calculate</u>:
- 23.640 g Mn(ClO₄)₂·6H₂O ÷ 361.84 g/mol = 0.0653 mol Mn(ClO₄)₂·6H₂O = mol Mn(ClO₄)₂
Now we divide the moles by the volume, to <u>calculate molarity</u>:
- 200 mL⇒ 200/1000 = 0.200 L
- 0.0653 mol Mn(ClO₄)₂ / 0.200 L = 0.3267 M
Answer:
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Explanation:
I HOPE IT HELPED
Answer:
Mass of water produced is 22.86 g.
Explanation:
Given data:
Mass of hydrogen = 2.56 g
Mass of oxygen = 20.32 g
Mass of water = ?
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 20.32 g/ 32 g/mol
Number of moles = 0.635 mol
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 2.56 g/ 2 g/mol
Number of moles = 1.28 mol
Now we will compare the moles of water with oxygen and hydrogen.
O₂ : H₂O
1 : 2
0.635 ; 2×0.635 = 1.27
H₂ : H₂O
2 : 2
1.28 : 1.28
The number of moles of water produced by oxygen are less thus it will be limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 1.27 × 18 g/mol
Mass = 22.86 g
