Answer:
The concentration is 50,8 % w/v and radio strengths = 1,96.
Explanation:
Phenobarbital sodium is a medication that could treat insomnia, for example.
2,0 M of Phenobarbital sodium means 2 moles in 1L.
The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means 1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:
2 moles × 
= 508 g of Phenobarbital sodium.
1 L ×
= 1000 mL of solution
Thus, % w/v is:
× 100 = 50,8 % w/v
And radio strengths:
= 1,96. Thus, you have 1 g in 1,96 mL
I hope it helps!
Answer:
0.077 M
Explanation:
Data Given :
The concentration of half normal (NaCl) saline = 0.45g / 100 g
So,
Volume of Solution = 100 g = 100 mL
Volume of Solution in Liter = 100 mL / 1000
Volume of Solution = 0.1 L
molar mass of NaCl = 58.44 g/mol
Molarity:
Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M
Formula used for Molarity
M = moles of solute / Liter of solution . . . . . . . . . . (1)
Now to find number of moles of Nacl
no. of moles of NaCl = mass of NaCl / molar mass
no. of moles of NaCl = 0.45g / 58.44 g/mol
no. of moles of NaCl = 0.0077 g
Put values in the eq (1)
M = moles of solute / Liter of solution . . . . . . . . . . (1)
M = 0.0077 g / 0.1 L
M = 0.077 M
So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M
Answer:
You will get 5.0 g of hydrogen.
Explanation:
As with any stoichiometry problem, we start with the balanced equation.
Sn
l
+
2HF
→
SnF
2
+
H
2
Moles of H
2
=
2.5
mol Sn
×
1 mol H
2
1
mol Sn
=
2.5 mol H
2
Mass of H
2
=
2.5
mol H
2
×
2.016 g H
2
1
mol H
2
=
5.0 g H
2
Explanation:
Dichloromethane is flammable - FALSE
Methanol is flammable. - TRUE
Concentrated sulfuric acid is corrosive. - TRUE
10% sodium carbonate solution must be used in the fume hood. - FALSE
Benzoyl chloride is a lachrymator. - TRUE
