Answer:
Mass = 55.52 g
Explanation:
Given data:
Number of atoms of Li = 4.81×10²⁴ atom
Number of grams = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
For Li:
4.81×10²⁴ atom × 1 mol / 6.022 × 10²³ atom
8 moles
Mass in gram:
Mass = number of moles × molar mass
Mass = 8 mol × 6.94 g/mol
Mass = 55.52 g
Answer:
0.571 mol
Explanation:
Given data:
Number of moles of NaHCO₃ = 0.571 mol
Number of moles of CO₂ produced = ?
Solution:
Chemical equation:
NaHCO₃ + C₃H₆O₃ → CO₂ + C₃H₅NaO₃ + H₂O
Now we will compare the moles of CO₂ with NaHCO₃ from balance chemical equation.
NaHCO₃ : CO₂
1 : 1
0.571 : 0.571
So number of moles of CO₂ produced are 0.571.
Answer:
half life=16.5hrs
sample=2g
total time=1day=24hrs
no. of half lives=total time/half life
no. of half lives=24hrs/16.5hrs=1.45approx1.5
sample left=1/2n=1/2[1.5]=0.707gapprox
Explanation:
During cellular respiration<span>, glucose is broken down in the presence of oxygen to produce carbon dioxide and water,energy is </span><span>released.</span>
Answer:
None of these
Explanation:
Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.
Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :
Step -1 : Generation of stable carbocation.
Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.
Step-2: Attack of the ring to the carbocation
The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.
The product formed is shown in mechanism does not mention in any of the options.
So, None of these is the answer
