2AgNO3 (aq) + Na2CO3 (aq) --> Ag2CO3 (s) + 2NaNO3 (aq)
2Ag(+) (aq) + CO3(2-) (aq) --> Ag2CO3 (s)
hope it helps
Answer:
There is an extra O2 molecule left over
Explanation:
It is effected by diffusion (the power of smell and wind spread) but a solid is not.
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
Answer:
Explanation:
Iso-electronic species have same number of electrons . Positive charged ions will have smaller size . As electrons add , size increases due to electronic repulsion .
Following species are isoelectronic .
Al³⁺ < Mg²⁺ < Na¹⁺ < Ne < F⁻¹ < O⁻² < N⁻³
