Answer:
Wt. Avg. Atomic Weight => 63.35457 amu
Explanation:
Given Isotopic %Abundance fractional Wt Avg
At. Mass (amu) abundance contribution
Cu-63 62.93 69.09 0.6909 43.4783
Cu-65 64.9278 20.0668 0.200668 20.0668
Wt Average of all isotopes = ∑Wt Avg Contributions
= 43.4783 amu + 20.0668 amu = 63.35457 amu
Answer:
9.79740949850 moles
Explanation:
- 1 mole = Avogardo's Number <<6.022 E 23 <<particles, atoms, etc.>>
- This problem can be solved using dimensional analysis by multiplying atoms (5.9E24 atoms) by (1) mole and then dividing the number by Avogardo's number (6.022 E 23 atoms).
- Note: E = * 10
Side Note: Please let me know if you need any clarifications about this!
One molecule of ammonia is composed of two atoms of nitrogen and three atoms of hydrogen. Option B.
<h3>What is an equation?</h3>
The term chemical equation has to do with the presentation of a chemical reaction on paper in a way that it can be easily understood. It is easy to write an equation to show what is going on in a reaction system.
Now we have the reactions as shown in the question. In this reaction which is the synthesis of ammonia and occurs industrially in the Haber process. The statement that is not true is that; one molecule of ammonia is composed of two atoms of nitrogen and three atoms of hydrogen. Option B.
Learn more about chemical equation:brainly.com/question/28294176
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Answer:
CH2O
Explanation:
Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.
C = 5.692/14.229 * 100 = 40%
O = 7.582/14.229 * 100 = 53.29%
H = 0.955/14.229 * 100 = 6.71%
We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.
C = 40/12 = 3.333
O = 53.29/16 = 3.33
H = 6.71/2 = 6.71
Dividing by the smaller value which is 3.33
C = 3.33/3.33 = 1
O = 3.33/3.33= 1
H = 6.71/3.33 = 2
The empirical formula of the compound ribose is CH2O