The molar mass and the mass of the substance give the moles of the substance. The 10-gram sample of neon (Ne) has the fewest atoms. Thus, option b is correct.
<h3>What are moles?</h3>
Moles of the substance are the ratio of the molar mass and the mass of the substance. The moles of each atom with respect to the Avagadro's number (Nₐ) is given as,
Moles = mass ÷ molar mass
- Moles of carbon (C) = 10 ÷ 12
= 0.83 Nₐ atoms
- Moles of neon (Ne) = 10 ÷ 20
= 0.5 Nₐ atoms
- Moles of fluorine (F) = 10 ÷ 19
= 0.52 Nₐ atoms
- Moles of nitrogen (N) = 10 ÷ 14
= 0.7 Nₐ atoms
- Moles of oxygen (O) = 10 ÷ 16
= 0.63 Nₐ atoms
Therefore, option b. 10 gm neon has the fewest atoms.
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Answer:
1.008moles of iodine
Explanation:
Hello,
This question requires us to calculate the theoretical yield of I₂ or number of moles that reacted.
Percent yield = (actual yield / estimated yield) × 100
Actual yield = 1.2moles
Estimated yield = ?
Percentage yield = 84%
84 / 100 = 1.2 / x
Cross multiply and solve for x
100x = 84 × 1.2
100x = 100.8
x = 100.8/100
x = 1.008moles
1.008 moles of I₂ reacted in excess of H₂ to give 1.2 moles of HI
Answer:
C
Explanation:
radiation from the sun first warms the outer atmosphere (trophosphere)
convection(aka just heat moving through gas or liquid) brings the warmth down lower
conduction heats the ground
gl lol :))
Answer:
1) acetylide
2) enol
3) aldehydes
4) tautomers
5) alkynes
6) Hydroboration
7) Keto
8) methyl ketones
Explanation:
Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.
Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.
The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.