Answer:
The correct answer is 0.033 M
Explanation:
We have a solution of NaClO with a concentration of 5%w/w:
5% w/w= 5 g NaClO/100 g solution
The first dilution is 10 ml of solution in 100 ml. That is a 1/10 dilution (10ml/100 ml= 1/10). That means we are diluting 10 times the solution. We can calculate the resulting concentration after this first dilution as follows:
5%w/w x 10 ml/100 ml = 5% w/w/10= 0.5%w/w
Then, we take 6 ml of 0.5% w/w solution and we add 6 ml of dye in a reaction vessel. The total volume of the solution in the reaction vessel is 6 ml + 6 ml= 12 ml, and we are diluting twice the solution because 6 ml/12 ml= 1/2. We can calculate the resulting concentration of the solution after this second dilution as follows:
0.5% w/w x 6 ml/12 ml= 0.5% w/w/2= 0.25%w/w
Finally, we need to convert the concentration from %w/w to M (mol solution/1L solution). For this, we assume a density of the solution close to the density of water (1.00 g/ml) and we use the molecular weight of NaClO (74.44 g/mol):
0.25 g NaClO/100 g solution x 1 mol NaClO/74.44 g x 1.00 g solution/1 ml x 100 ml/0.1 L= 0.033 mol/L
= 0.033 M
Answer:
Explanation:
To calculate the reduced mass of nitrogen molecule = Mn x Mn/ Mn + Mn
Mn = atomic mass of Nitrogen = 14.00
reduced mass = 14 x 14 /14 + 14 = 7
Similarly for HCl; reduced mass for a hydrogen chloride molecule = Mh x Mcl/Mh + Mcl
where Mh = atomic mass of hydrogen = 1.008
Mcl = atomic mass of chrorine = 34.97
reduced mass for HCl = 1.008 x 34.97 /1.008 + 34.97
= 0.97
Answer:
d) n = 5.4 mol
Explanation:
Given data:
Number of moles of He = ?
Volume of He gas = 120 L
Temperature and pressure = standard
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
1 atm × 120 L = n × 0.0821 atm.L/ mol.K × 273.15 K
120 atm.L = n × 22.43 atm.L/ mol
n = 120 atm.L / 22.43 atm.L/ mol
n = 5.4 mol
A, b and d if you don't count the tooth.
