Answer:
C. Mutations are a change in DNA or a chromosome and can be helpful, harmful or may have no affect.
Explanation:
- Mutations are spontaneous random changes that occurs in the genetic make up of an organisms. Mutations are rare and their rate of occurrence is random.
- Mutations may occur on the gene level known as gene mutations or at chromosome levels called chromosomal mutations.
- Mutations may be beneficial, harmful or have no effect on a given organisms. Harmful mutations cause disorders that may lead to abnormality or death of an organisms. Beneficial mutations improve an organisms adaptability to the environment.
Answer:
Natural chemicals are produced by nature without any human intervention. Synthetic chemicals are made by humans using methods different than those nature uses, and these chemical structures may or may not be found in nature. Stay safe .
Given parameters:
Volume of CuSO₄ = 250mL
Concentration of CuSO₄ = 2.01M
Unknown:
Mass of CuSO₄.5H₂O = ?
To solve this problem, we must write the chemical relationship between both species.;
CuSO₄.5H₂O → CuSO₄ + 5H₂O
Now that we know the expression, it is possible to solve for the unknown mass.
First find the number of moles of CuSO₄;
Number of moles = Concentration x Volume
Take 250mL to L so as to ensure uniformity of units;
Volume = 250 x 10⁻³L
Input the parameters and solve for number of moles;
Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol
From the equation;
1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O
So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O
Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole
Mass of CuSO₄.5H₂O = number of moles x molar mass
= 0.5 x 249.6
= 124.8g
The mass of CuSO₄.5H₂O is 124.8g
Answer:
Explanation:
2C₂H₅OH = C₄H₆ + 2H₂O + H₂
2 mole 1 mole
molecular weight of ethyl alcohol
mol weight of C₂H₅OH = 46 gm
mol weight of C₄H₆ 54 gm
540 gm of C₄H₆ = 10 mole
10 mole of C₄H₆ will require 20 mol of ethyl alcohol .
20 mole of ethyl alcohol = 20 x 46
= 920 gm
ethyl alcohol required = 920 gm .
