Answer:
3.56L will be the obtained volume
Explanation:
0.3 M → Molarity
These are the moles of solute (for this case, the solute is KOH) that are contianed in 1L of solution
We dissolved 60 g of solute, so let's determine the moles:
60 g . 1 mol/ 56.1 g = 1.07 moles
Molarity(mol/L) = Mol / Volume(L)
Let's replace data: 0.3M = 1.07 mol / Volume(L)
0.3 mol/L = 1.07 mol / volume
Volume = 1.07 mol / 0.3mol/L = 3.56 L
(3) from 400.oC to 200.oC
(4) from 200.oC to 400.oC
Spell out the full name of the compound.
Answer:
We do not have - the amount of substance.
The molecular weight () of Bromine is 80.
Answer: The mass of excess reagent left is 20.48 g
Explanation:
To calculate the number of moles, we use the equation:
.....(1)
Given mass of sulfur = 50 g
Molar mass of sulfur = 32 g/mol
Putting values in equation 1, we get:
Given mass of fluorine gas = 105 g
Molar mass of fluorine gas = 37.99 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of sulfur and fluorine gas follows:
By Stoichiometry of the reaction:
3 moles of fluorine gas reacts with 1 mole of sulfur
So, 2.76 moles of fluorine gas will react with = of sulfur
As, given amount of sulfur is more than the required amount. So, it is considered as an excess reagent.
Thus, fluorine gas is considered as a limiting reagent because it limits the formation of product.
Moles of excess reagent (sulfur) left = 1.56 - 0.92 = 0.64 moles
Molar mass of sulfur = 32 g/mol
Moles of sulfur = 0.64 moles
Putting values in equation 1, we get:
Hence, the mass of excess reagent left is 20.48 g