Answer:
5.01%
Explanation:
Density of vinegar = mass/volume
Mass of 10.00 mL = density x volume
= 1.006 x 10 = 10.06 g
From the equation of reaction:
1 mole pf CH3COOH requires 1 mole of NaOH for neutralization.
mole of NaOH = molarity x volume
= 0.5062 x 0.01658
= 0.008392796 mole
0.008392796 mole of NaOH will therefore require 0.008392796 mole of CH3COOH.
mass of CH3COOH = mole x molar mass
= 0.008392796 x 60.052
= 0.504 g
Percentage by mass of acetic acid in the vinegar = 0.504/10.06 x 100%
= 5.01%
The percent by mass of acetic acid in the vinegar is 5.01%
Firstly the limiting reactant should be identified. Limiting reactant is the reactant that is in limited supply, the amount of product formed depends on the moles present of the limiting reactant.
the stoichiometry of x to y = 1:2
1 mole of x reacts with 2 moles of y
if x is the limiting reactant, there are 3 moles of x, then 6 moles of y should react, however there are only 4 moles of y. Therefore y is the limiting reactant and x is in excess.
4 moles of y reacts with 2 moles of x
since there are 3 moles of x initially and only 2 moles are used up, excess amount of x is 1 mol thats in excess.
Answer:
will be 90054 J
Explanation:
Number of moles = (mass)/(molar mass)
Molar mass of = 134.45 g/mol
So, 1.00 g of = of = 0.00744 mol of
0.00744 mol of produces 670 J of heat
So, 1 mol of produces of heat or 90054 J of heat
Answer:
Mass = 1.33 g
Explanation:
Given data:
Mass of argon required = ?
Volume of bulb = 0.745 L
Temperature and pressure = standard
Solution:
We will calculate the number of moles of argon first.
Formula:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
By putting values,
1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K
0.745 atm. L = n × 22.43 atm.L/mol
n = 0.745 atm. L / 22.43 atm.L/mol
n = 0.0332 mol
Mass of argon:
Mass = number of moles × molar mass
Mass = 0.0332 mol × 39.95 g/mol
Mass = 1.33 g
Answer:
if you want to find average speed so u need to use this formula
Average speed =