Answer:
pH = 12.65
Explanation:
From the given information:
number of moles =mass in gram / molar mass
number of moles of KOH = mass of KOH / molar mass of KOH
number of moles of KOH = 0.251 g / 56.1 g/mol = 0.004474 mol
For solution :
number of moles = Concentration × volume
concetration = number of moles/ volume
concetration = 0.004474 mol / 0.100 L
concetration = 0.04474 M
We know that 1 moles KOH result into 1 mole OH⁻ ions
Therefore, Molarity of OH⁻ = 0.04474 M
Now,
pOH = -log[OH⁻]
pOH = -log (0.04474) M
pOH = 1.35
Similarly,
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 1.35
pH = 12.65
Answer:
A particle consisting of two neutrons and two protons, with a positive charge; emitted energetically from the nuclei of unstable isotopes of mass number 82 and up. Synonym(s): alpha ray. Helie, Louis T.
The outcome of the equation shows that there is no light energy proving that light energy was absorbed to get CH20; +602
Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g
They increase across each period, decrease down a group. As you go across a period the number of protons and increases. The positive nucleus then has a stronger attractive force on the electrons so it takes a larger amount of energy to remove an electron. As you go down a group the atoms are larger so the attractive force is weaker and it takes less energy to remove an electron.
