[H+] in first brand:
4.5 = -log([H+])
[H+] = 10^(-4.5)
[H+] in second brand:
5 = -log[H+]
[H+] = 10^(-5)
Difference = 10^(-4.5) - 10^(-5)
= 2.2 x 10⁻⁵
The answer is A.
Check attached file for the answer.
Answer:
Upper F subscript 2 (g) plus upper C a (s) right arrow with delta above upper C a upper F subscript 2 (s).
Explanation:
This is a chemical reaction problem.
In expressing any chemical reaction, we need to understand that there are reactants and products.
- The reactants are the species on the left hand side that are combining.
- The products are the species on the right hand side that are formed.
- Every chemical reaction is obeys the law of conservation of matter i.e equal number of matter on both sides.
Using the statement of this problem, we can deduce that;
Reactants are Fluorine gas and Calcium metal
Product is Calcium Fluoride
Note: A metal is a solid(s) and powder is a solid(s). A gas is denoted as (g). They depict the state of the species reacting.
F₂
+ Ca
→ CaF₂
We can see that equal number of atoms are on both sides of the expression.
Answer: 1.Stars are born in clouds of gas and dust called nebulas.
2.The gas and dust are pulled together by gravity.
3.Heat and pressure cause nuclear fusion, which signals the birth of a star.
Explanation:
Answer:
SO2(g) and H2S(g)
Explanation:
Sulphuric acid (H2SO4) is very important in industries and is often referred to as the best chemical due to its multiple uses. It is corrosive and used in making fertilizers and explosives. It can also be used to make paints, detergents etc.
Sulphuric acid reacts with Sodium iodide to give Hydrogen iodide and Sodium hydrogen Sulphate.
NaI (s) + H2SO4 (l) =HI (g) + NaHSO4 (s)
The Hydrogen iodide then reacts with the Sulphuric acid to give Iodine,Hydrogen sulphide and water
HI (g) + H2SO4 (l) = 4I2(s) + H2S (g) + 4 H2O (l)
The sulphuric acid can also undergo reduction reaction with hydrogen iodide to give Iodine, Sulphur dioxide and Water.
2 HI (g) + H2SO4 (l) = I2(s) + SO2 (g) + H2O (l)
