Mole ratio:
MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl
2 moles KOH ---------------- 1 mole Mg(OH)₂
4 moles KOH ------------------- moles Mg(OH)₂
moles Mg(OH)₂ = 4 x 1 / 2
= 2 moles of Mg(OH)₂
molar mass Mg(OH)₂ = 58g/mol
mass of Mg(OH)₂ = n x mm
mass of Mg(OH)₂ = 58 x 2
= 116 g of <span> Mg(OH)₂</span>
hope this helps!
The Volumes can be calculated from Masses by using following Formula,
Density = Mass / Volume
Solving for Volume,
Volume = Mass / Density
Mass of Both Gases = 14.1 g
Density of Argon at S.T.P = 1.784 g/L
Density of Helium at S.T.P = 0.179 g/L
For Argon:
Volume = 14.1 g / 1.784 g/L
Volume = 7.90 L
For Helium:
Volume = 14.1 g / 0.179 g/L
Volume = 78.77 L
Answer:
Mass percent of food dyes = 0.0616%
Explanation:
Given data:
Mass of candy = 47.9 g
Calories = 240
Mass of fat = 10 g
Mass of carbohydrate = 34 g
Mass of protein = 2 g
Mass of food dyes = 29.5 mg
Mass percent of food dyes = ?
Solution:
First of all we will convert the mg into g.
Mass of food dyes = 29.5 mg × 1g /1000 mg = 0.0295 g
Mass percent of food dyes = mass of food dyes / total mass× 100
Now we will put the values.
Mass percent of food dyes = 0.0295 g / 47.9 g × 100
Mass percent of food dyes = 0.000616 × 100
Mass percent of food dyes = 0.0616%
Answer:
2.52atm for sure
Explanation:
We know that:
P1= 2.80atm
T1= 400k
T2= 360k
And we want to find P2= ?
We have an equation: P1/T1 = P2/T2
Multiple both side by T2 to remove it
P2 = 2.52