Answer:
The answer to your question is 3.69 x 10²¹ atoms of gold
Explanation:
Data
density = 19.32 g/cm³
dimensions = 2.5 cm x 2.5 cm x 0.1 mm
number of atoms of gold = x
Process
1.- Find the volume of the foil
Volume = length x height x width
Volume = 2.5 x 2.5 x 0.01
Volume = 0.0625 cm³
2.- Calculate the mass of the foil
Density = mass/volume
mass = density x volume
mass = 19.32 x 0.0625
mass = 1.208 g
3.- Calculate the number of atoms
Atomic number of Gold = 197 g
197 g -------------------- 6.023 x 10²³ atoms
1.208 g --------------- x
x = (1.208 x 6.023 x 10²³)/197
x = 3.69 x 10²¹ atoms of gold
though alchemists were often superstitious, they left a rich legacy of modern chemists. what was their main contribution-
Explanation:
they were the first to preform experiments.
Answer:
0.0014 moles is present in 40cm³ of 0.035M of HCl solution
Explanation:
Molarity = 0.035M
V = 40.0mL
1mL = 1cm³
V = 40cm³
0.035 moles = 1000cm³
X moles is present in 40cm³
X = (40 * 0.035) / 1000
X = 0.0014moles
0.0014 moles is present in 40cm³ of solution
The greatest aqueous freezing point is (D) 0.10 KCI
