Molarity is expressed as
the number of moles of solute per volume of the solution. For example, we are
given a solution of 2M NaOH this describes a solution that has 2 moles of NaOH
per 1 L volume of the solution. We calculate as follows:
0.115 M = n mol KBr / .55 L solution
n = 0.06325 mol KBr
mass = 0.06325 mol KBr (119 g / mol) = 7.53 g KBr
D has a total of four significant figures.
Answer:
The correct order will be
a. Transfer the measured amount of NaCl to the volumetric flask.
e. Dissolve the NaCl in less than 250 mL of water and mix well.
b. Dilute the solution with water to the 250.0 mL mark.
Explanation:
Preparation of NaCl solution in 250.0 ml volumetric flask:
Add the weighed NaCl directly to volumetric flask and add small amount of water to it and mix it will until all NaCl gets dissolved( if not add small water amount of water more)
After dissolving NaCl add the water upto the mark.
The correct order will be
a. Transfer the measured amount of NaCl to the volumetric flask.
e. Dissolve the NaCl in less than 250 mL of water and mix well.
b. Dilute the solution with water to the 250.0 mL mark.
Answer is: directly up.
Neon (Ne) is noble gas with atomic number 10 and atomic mass around 20.
1) directly up is noble gas helium (He) with atomic mass around 4.
2) directly down is noble gas argon (Ar) with atomic mass around 40.
3) directly to the left is florine (F), with smaller atomic mass, but also from different group.
4) directly to the right there no elements, because noble gases are far right in 18. group of Periodic table.
Answer: 91.73g of NaCl
Explanation:
First, we solve for the number of moles of F2 using the ideal gas equation
V = 12L
P = 1.5 atm
T = 280K
R = 0.082atm.L/mol/K
n =?
PV = nRT
n = PV /RT
n = (1.5x12)/(0.082x280)
n = 0.784mol
Next, we convert this mole ( i.e 0.784mol) of F2 to mass
MM of F2 = 19x2 = 38g/mol
Mass conc of F2 = n x MM
= 0.784 x 38 = 29.792g
Equation for the reaction is given below
F2 + 2NaCl —> 2NaF + Cl2
Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol
Mass conc. of NaCl from the equation = 2 x 58.5 = 117g
Next, we find the mass of NaCl that reacted with 29.792g of F2.
From the equation,
38g of F2 redacted with 117g of NaCl.
Therefore, 29.792g of F2 will react with Xg of NaCl i.e
Xg of NaCl = (29.792 x 117)/38
= 91.73g
Therefore, 91.73g of NaCl reacted with f2