The chemical formula for the product is RbBr.
The mole fraction of methanol in the mixture is 0.444
We'll begin by calculating the number of mole of water.
- Molar mass of water = 18 g/mol
Mole = mass / molar mass
Mole of water = 45 / 18
Mole of water = 2.5 moles
Finally, we shall determine the mole fraction of methanol.
- Mole of water = 2.5 moles
- Mole of methanol = 2 moles
- Total mole = 2 + 2.5 = 4.5 moles
Mole fraction of methanol =?
Mole fraction = mole / total mole
Mole fraction of methanol = 2 / 4.5
Mole fraction of methanol = 0.444
Thus, the mole fraction of methanol is 0.444
Learn more about mole fraction:
brainly.com/question/15444997
This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:
Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰
Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻
Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.
In such a way, the correct choice is C.
Learn more:
Answer:
A
Explanation:
Mass is a raw measurement, weight is effected by gravity
I’m assuming you mean barium nitrite, Ba(NO2)2.
First convert grams of Ba(NO2)2 to moles using the molar mass of Ba(NO2)2. Then use the mole ratio of 4 moles of oxygen per 1 mole of Ba(NO2)2 to convert to moles of oxygen. Then use the molar mass of oxygen to convert to grams of oxygen.
45.7 g Ba(NO2)2 • 1 mol Ba(NO2)2 / 229.35 g Ba(NO2)2 • 4 mol O / 1 mol Ba(NO2)2 • 16.0 g O / 1 mol O = 12.8 g oxygen
