Answer:
The chlorine gas and potassium bromide solution react to form liquid bromine and potassium chloride solution.
Explanation:
Chemical equation:
Cl₂(g) + KBr (aq) → KCl (aq) + Br₂(l)
Balanced chemical equation:
Cl₂(g) + 2KBr (aq) → 2KCl (aq) + Br₂(l)
This equation showed that the chlorine gas and potassium bromide solution react to form liquid bromine and potassium chloride solution.
Chlorine is more reactive than bromine it displace the bromine from potassium and form potassium chloride solution.
The given equation is balanced and completely hold the law of conservation of mass.
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
Explanation:
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Every substance is either an element or a compound.
There is ionic compounds, molecular compounds, but they are all considered compounds.
Hope I helped.
Answer:
I believe Si12H26+02 is the coefficient I might be wrong, Sorry if I am.
-Photons are absorbed by hot gas atoms
-Energy is transferred through large-scale movement of material
-Energy is released into the photosphere
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
