I will have to say that this statement is true
Answer:
σ = 4.998 E-4 C/m²
Explanation:
- 1 Coulomb (C) ≡ 6.241509 E18 electrons (e)
∴ # elect = 6.24 E14 elect
charge (Q):
⇒ Q = (6.24 E14 elect)/( 1 C /6.241509 E18 elect) = 9.998 E-5 C
charge density (σ):
∴ surface area (S) = 0.2 m²
⇒ σ = ( 9.998 E-5 C ) / ( 0.2 m²)
⇒ σ = 4.998 E-4 C/m²
To get the percent yield, we will use this formula:
((Actual Yield)/(Theoretical Yield)) * 100%
Values given: actual yield is 220.0 g
theoretical yield is 275.6 g
Now, let us substitute the values given.
(220.0 grams)/(275.6 grams) = 0.7983
Then, to get the percentage, multiply the quotient by 100.
0.7983 (100) = 79.83%
Among the choices, the most plausible answer is 79.8%
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Answer:
2.01 moles of P → 1.21×10²⁴ atoms
2.01 moles of N → 1.21×10²⁴ atoms
4.02 moles of Br → 2.42×10²⁴ atoms
Explanation:
We begin from this relation:
1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br
Then 2.01 moles of PNBr₂ will have:
2.01 moles of P
2.01 moles of N
4.02 moles of Br
To determine the number of atoms, we use the relation:
1 mol has NA (6.02×10²³) atoms
Then: 2.01 moles of P will have (2.01 . NA) = 1.21×10²⁴ atoms
2.01 moles of N (2.01 . NA) = 1.21×10²⁴ atoms
4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms
