The correct answer is 3) 2CO2(g) ⇄ 2CO(g) + O2(g)
this is the correct one because it is a decomposition reaction and all the number of atoms is equal on both sides.
there are 2 C atoms on both sides.
and 4 O atoms on both sides.
and 1) the atoms numbers are equal on both sides but not correct as it not a
correct number as it has 1/2 O2.
and 2) CO2(g) ⇆ CO(g) + O2
the number of O atoms is not equal on both sides of the equation.
we have 2 O atoms on the left side and 3 O atoms on the right side.
so, this not a balanced equation.
4) also not correct 2CO(g) + O2 ⇆ 2CO2
as it is not a decomposition reaction and the 2CO & O2 are as reactants not products.
so the correct answer is 3) 2CO2(g) ⇆ 2CO(g) + O2(g)
Answer: Limitation: They may be more expensive and time consuming than lab experiments. Limitation: There is no control over extraneous variables that might bias the results. This makes it difficult for another researcher to replicate the study in exactly the same way.
Explanation:
Hope this helps!
Answer:
See explanation
Explanation:
The molecular equation shows all the compounds involved in the reaction.
The molecular equation is as follows;
2NaF(aq) + Pb(NO3)2(aq) -------> PbF2(s) + 2NaNO3(aq)
The complete ionic equation shows all the ions involved in the reaction
The complete ionic equation;
2Na^+(aq) + 2F^-(aq) + Pb^2+(aq) + 2NO3^-(aq) -------->PbF(s) + 2Na^+(aq) +2NO3^-(aq)
The net Ionic equation shows the ions that actually participated in the reaction
The net ionic equation is;
2F^-(aq) + Pb^2+(aq)--------> PbF(s)
The color of light that will be diffracted at a greater angle from a diffracting grating is THE YELLOW COLOR.
This is because, the degree of diffraction depend on the wavelength of light and light color with shorter wavelength are diffracted at a larger angle than those with longer wavelengths. The wavelength of the yellow color is much more smaller than that of the red color.
Carbonated drinks are a mixture of many compounds such as glucose(sugar) and others, while water is simply its own molecule.
