Answer: D
Explanation:
Answer: High temperatures and location in high latitudes
Explanation:
Answer: The molecule identified with the letter C. HFO.
Explanation:
1) In covalent bonds, the dipolesare identified with an arrow pointing from the least electronegative element toward the most electronegative element.
2) The molecule HFO is formed by one covalent bond between H and O, and other covalent bond between F and O.
3) O is more electronegative than H, so in the H-O bond the dipole is identified with the arrow from the H atom toward the O atom. That is correctly indicated in the drawing C, in this way:
-|----→
H - O
4) F is more electronegative than O, so in the O-F bond the dipole is identified with an arrow from the O atom toward the F atom. That is correctly indicated in this way:
-|---→
O - F.
b. Oxygen
c. Hydrogen
d. Nitrogen
Answer: The correct answer is Option d.
Explanation:
Dry air is composed of a mixture of gases like argon, oxygen, nitrogen, carbon dioxide and many other gases present in trace amount.
The percentage of these gases by volume in dry air is:
Carbon dioxide = 0.03%
Argon = 0.93%
Oxygen = 20.95%
Nitrogen = 78.09%
Hydrogen = Trace amounts
From the above percentage values, we can easily say that nitrogen has the highest percentage by volume in dry air. Hence, the correct option is d.
In the given question, in 28 grams of molecule 0.636 moles are present.
A mole is the amount of a substance which comprises exactly 'elementary entities' of that substance.
To find the number of moles in 28 grams of , we first need to determine the molar mass of .
Molar mass of = 12.01 g/mol (carbon) + 2(16.00 g/mol) (oxygen)
= 44.01 g/mol
Now that we know the molar mass of , we can use the following formula to calculate the number of moles:
= 0.636 mol
Therefore, there are 0.636 moles in 28 grams of .
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Given:
Molar mass CO2 is 44.01 g/mol
25.5 g of CO2
Required:
Moles of LiOH
Solution:
Balanced equation is:
2LiOH +CO2 → Li2CO3 + H2O
25.5g CO2 (1 mole CO2/44.01 g/mol CO2)(2 moles LiOH/1 mol CO2) = 1.16 moles LiOH
Molarmass carbon dioxide is 44.01 g/mol, 1.16 moles of LiOH are needed to react completely with 25.5 g of CO₂.
Stoichiometry and the molar ratios of CO₂ and LiOH must be used to calculate the number of moles of LiOH required to totally react with 25.5 g of CO₂.
The reaction between CO₂ and LiOH has the following chemical formula:
CO₂ + 2LiOH -> Li₂CO₃ + H₂O
Now,
Molar mass of CO₂ = 44.01 g/mol
Mass of CO₂ = 25.5 g
Number of moles of CO₂ = Mass of CO2 / Molar mass of CO2
= 25.5 g / 44.01 g/mol
≈ 0.58 moles of CO2
As per this,
Number of moles of LiOH = 2 * Number of moles of CO2
= 2 * 0.58 moles
= 1.16 moles of LiOH
Thus, 1.16 moles of LiOH are needed for the given reaction.
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