Answer:
it is a pathways by which a chemical substance moves thru biotic and abiotic components of earth. example is the water cycle
1 gram = 1000 mg
=>
1 mg = 1/1000 g
25 mg = 25* (1/1000) g = 25/1000 g = 0.025g
Answer:
Average atomic mass = 79.9035 amu.
The given element is bromine.
Explanation:
Given data:
Mass of 1st isotope = 78.9183 amu
Percent abundance of 1st isotope = 50.69%
Mass of 2nd isotope = 80.9163 amu
Percent abundance of 1st isotope = 49.31%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50.69×78.9183 )+(49.31× 80.9163) /100
Average atomic mass = 4000.3686+ 3989.9828 / 100
Average atomic mass = 7990.3514 / 100
Average atomic mass = 79.9035 amu.
The given element is bromine.
The volume of methanol necessary to prepare the antifreeze good for antifreeze solution is 3.2 L
<h3>Dilution formula</h3>
M₁V₁ = M₂V₂
Where
- M₁ is the molarity of stock solution
- V₁ is the volume of stock solution
- M₂ is the molarity of diluted solution
- V₂ is the volume of diluted solution
<h3>Data obtained from the question </h3>
- Molarity of stock solution (M₁) = 24.7 mol/L
- Volume of diluted solution (V₂) = 8 L
- Molarity of diluted solution (M₂) = 10 mol/L
- Volume of stock solution needed (V₁) = ?
<h3>How to determine the volume needed </h3>
The volume of the methanol necessary to prepare the solution can be obtained as illustrated below:
M₁V₁ = M₂V₂
24.7 × V₁ = 10 × 8
24.7 × V₁= 80
Divide both side by 24.7
V₁ = 80 / 24.7
V₁ = 3.2 L
Thus, the volume of methanol necessary to prepare the antifreeze good for antifreeze solution is 3.2 L
Learn more about dilution:
brainly.com/question/15022582
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