Answer is: there is 2,69·10²³ atoms of bromine.
m(CH₂Br₂) = 39,0 g.
n(CH₂Br₂) = m(CH₂Br₂) ÷ M(CH₂Br₂).
n(CH₂Br₂) = 39 g ÷ 173,83 g/mol.
n(CH₂Br₂) = 0,224 mol.
In one molecule of CH₂Br₂, there is two bromine atoms, so:
n(CH₂Br₂) : n(Br) = 1 : 2.
n(Br) = 0,448 mol.
N(Br) = n(Br) · Na.
N(Br) = 0,448 mol · 6,022·10²³ 1/mol.
n(Br) = 2,69·10²³.
The correct answer is: [C]: " mg " {"milligrams"} .
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I would think it is a heterogeneous mixture since it can't be an element since there are more than one type of atom, it can't be a compound since the leaves are not bonded together, and it can not be a homogeneous mixture since the leaves don't all blended together (the pile is not uniform) and you can distinguish all the different parts of the mixture. It can be considered a heterogeneous mixture since the leaves are mixed together (along with other things like dirt) in a non-uniform way so that you can point out the parts of the mixture and it does not look like one thing.
I hope this helps. Let me know in the comments if anything is unclear.
Data Given:
Time = t = 30.6 s
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 63.54/2 = 31.77 g
Amount Deposited = W = ?
Solution:
According to Faraday's Law,
W = I t e / F
Putting Values,
W = (10 A × 30.6 s × 31.77 g) ÷ 96500
W = 0.100 g
Result:
0.100 g of Cu²⁺ is deposited.
3 mol H₂ → 2 mol NH₃
5 mol H₂ → x mol NH₃
x=2*5.0/3=3.3
n(NH₃)=3.3 mol
