Answer:
Explanation:
First, we need to find the molecular mass of water (H₂O).
H₂O has:
- 2 Hydrogen atoms (subscript of 2)
- 1 Oxygen atom (implied subscript of 1)
Use the Periodic Table to find the mass of hydrogen and oxygen. Then, multiply by the number of atoms of the element.
- Hydrogen: 1.0079 g/mol
- Oxygen: 15.9994 g/mol
There are 2 hydrogen atoms, so multiply the mass by 2.
- 2 Hydrogen: (1.0079 g/mol)(2)= 2.0158 g/mol
Now, find the mass of H₂O. Add the mass of 2 hydrogen atoms and 1 oxygen atom.
- 2.0158 g/mol + 15.9994 g/mol = 18.0152 g/mol
Next, find the amount of moles using the molecular mass we just calculated. Set up a ratio.
Multiply. The grams of H₂O will cancel out.
The original measurement given had two significant figures (3,2). We must round to have 2 significant figures. All the zeroes before the 1 are not significant. So, round to the ten thousandth.
The 7 in the hundred thousandth place tells us to round up.
There are about <u>0.0018 moles in 0.032 grams.</u>
The correct answer is - A) The major constituents of air are gaseous elements.
With the statement ''the major constituents of air are gaseous elements'' we can easily conclude that the air is a mixture. The reason for that is that we have a plural usage of the word element, elements, which mean that there are multiple elements that make up the air.
The air is indeed predominantly a mixture of gaseous elements. The most abundant gas in the air being the nitrogen with 78.9%, oxygen with 20.95%, argon 0.93%, and carbon dioxide 0.04%, with lesser amounts of other gases also be present in it. The water vapor is also present in the air, though it is variable, being around 1% at sea level, but only 0.4% over the entire atmosphere.
The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.
Isotope mass amu Relative abundance
1 77.9 14.4
2 81.9 14.3
3 85.9 71.3
Express your answer to three significant figures and include the appropriate units.
Answer: 84.2 amu
Explanation:
Mass of isotope 1 = 77.9
% abundance of isotope 1 = 14.4% = 
Mass of isotope 2 = 81.9
% abundance of isotope 2 = 14.3% = 
Mass of isotope 3 = 85.9
% abundance of isotope 2 = 71.3% = 
Formula used for average atomic mass of an element :
![A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%2877.9%5Ctimes%200.144%29%2B%2881.9%5Ctimes%200.143%29%2B%2885.9%5Ctimes%200.713%29%5D)
Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu
Answer:
The answer of the scientific STATEMENT is" Appears That Ants Live In Colonies."
Answer:
Empirical evidence for a proposition is evidence, i.e. what supports or counters this proposition, that is constituted by or accessible to sense experience or experimental procedure. Empirical evidence is of central importance to the sciences and plays a role in various other fields, like epistemology and law.
Explanation:
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