I'm pretty sure it's called a volcanic winter, look up my answer to be correct though.
Answer:
6.4 g BaSO₄
Explanation:
You have been given the molarity and the volume of the solution. To find the mass of the solution, you need to (1) find the moles BaSO₄ (via the molarity ratio) and then (2) convert moles BaSO₄ to grams BaSO₄ (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given values.
Molarity (mol/L) = moles / volume (L)
(Step 1)
55 mL / 1,000 = 0.055 L
Molarity = moles / volume <----- Molarity ratio
0.5 (mol/L) = moles / 0.055 L <----- Insert values
0.0275 = moles <----- Multiply both sides by 0.055
(Step 2)
Molar Mass (BaSO₄): 137.33 g/mol + 32.065 g/mol + 4(15.998 g/mol)
Molar Mass (BaSO₄): 233.387 g/mol
0.0275 moles BaSO₄ 233.387 g
--------------------------------- x ------------------- = 6.4 g BaSO₄
1 mole
The volume of a substance is simply the ratio of mass and
density. Therefore:
volume = mass / density
Calculating for volume of Carbon Tetrachloride that the
student has to pour out:
volume = 55.0 g / (1.59 g / cm^3)
<span>volume = 34.60 cm^3</span>
D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
