Answer:
D. One dependent variable and one experimental variable
Explanation:
The experimental variable is defined as an independent variable which can be manipulated during an experiment in order to find out its impact or influence on the dependent variable.
The dependent variable refers to a variable that changes when an experimental or independent variable is manipulated.
Every experiment must have at least one experimental variable and one dependent variable in order to yield any meaningful result.
Chemical compounds have chemically bonded molecules so that they exhibit different properties (i.e. chemical) compared to the individual molecules comprising the compound. Mixtures are simply the combinations of different molecules and compounds that are not chemically bonded together, and can therefore be separated by physical means. Mixtures usually retain the properties of its components. The hydrogen and oxygen molecules in a mixture do not form strong bonds between each other. The molecules of both gases are only contained in the same space or volume and the individual molecules retain their chemical properties.
A compound containing hydrogen and oxygen molecules exhibit different chemical (and even physical) properties compared to the individual molecules themselves.
Water for example, is a compound with 2 hydrogen atoms and 1 oxygen atom, chemically-bonded together. Hydrogen gas is highly flammable, water is not. Oxygen gas is an essential reagent for combustion (or burning) reactions, water is not.
Thus, throwing a lighted match to a gaseous mixture of hydrogen

and oxygen

would create fire, or even an explosion (since hydrogen is flammable and oxygen feeds the reaction). Throwing a match to water vapor

would not create fire.
Answer:
Explanation:
use the equation
moles = mass/mr
=19.9/79.5
=0.250moles of CuO
then do the same for
H = 2.02/1
=2.02
so CuO is the limiting reagent because there is less amount of it.
Hope this helps :)
Answer:
The compound with the correct formula is;
A. MNO₃
Explanation:
The number of oxidation states in the metal, M = One oxidation state
The formula of the compound formed by the metal, M = MHCO₃
We note that the ion HCO₃⁻, known as hydrogen carbonate has an oxidation number of -1
Similarly nitrate, NO₃⁻ has an oxidation number of -1, therefore, the metal M can form similar compound formed with HCO₃⁻ with nitrate, and we have;
The possible compounds formed by the metal 'M' includes MHCO₃ and MNO₃.
Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v