The correct answers are
Free enterprise and government
:)
Answer:
Different visible waves show different colors
Answer:
Redox reaction and single displacement
Explanation:
This reaction is first of all a redox reaction. A redox reaction is a reaction that involves both oxidation and reduction. Oxidation involves increase in oxidation number while reduction involves decrease in oxidation number.
Copper (Cu) had an oxidation number of "0" as a reactant but had an oxidation number of "2+" in the product [Cu(NO₃)₂] hence oxidation occurred.
Nitrogen (N) had an oxidation number of "5+" in the reactant (HNO₃) but had an oxidation number of "4+" in the product (NO₂) hence reduction also occurred.
Also, from the reaction, it can be deduced that copper (Cu) displaced hydrogen (H) from the nitric acid (HNO₃) solution to form copper (II) nitrate [Cu(NO₃)₂]. It should be noted that copper can displace hydrogen because it is higher than hydrogen in the electrochemical series. Hence, this reaction can also be called a single displacement reaction. A single displacement reaction is a reaction in which an atom of an element replaces another atom in a compound (as seen in the equation given in the question).
Answer:
50 mL
Explanation:
We can solve this problem by using Boyle's Law, which states that:
"For a fixed mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume"
Mathematically:
where
p is the pressure of the gas
V is the volume of the gas
The equation can be rewritten as
where in this problem:
is the initial pressure of the gas in the coke
is the initial volume
is the final pressure
Solving for V2, we find the final volume:
The solubility (in M) of O2 in the blood of a scuba diver at a depth of 100 feet is 2.32 × 10^-3 mol.
<h3>How to calculate the solubility?</h3>
It should be noted that the partial pressure of oxygen will be:
= Mole fraction × Total pressure
= 0.209 × 3
= 0.627
According to Henry's law, the solubility will be:
= 3.7 × 10^-2 × 0.627
= 2.32 × 10^-3 mol
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