Which type of circuit has more than one path?
1 answer:
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H₂S
<h3>Further explanation</h3>Given
ΔH fusion and ΔH vaporization of different substances
Required
The substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes
Solution
We can use the formula :
Q=heat/energy absorbed
n = moles
The heat absorbed : 58.16 kJ
moles = 3.11
so ΔH vaporization :
The correct substance which has ΔH vaporization = 18.7 kj / mol is H₂S
(H₂S from the data above has ΔH fusion = 2.37 kj / mol and ΔH vaporization = 18.7 kj / mol)
A stable arrangement of eight valence electrons : ³⁵Cl⁻¹
<h3>Further explanation</h3>Chlorine is a halogen gas, located in group 17, p block
Chlorine has an atomic number of 17 and an atomic mass of 35
Electron configuration: [Ne] 3s²3p⁵
If we look at the electron configuration, then Cl will bind 1 more electron so that the configuration is stable like Argon (atomic number 18)
So by binding this one electron, chlorine forms negative ions (anions)
³⁵Cl⁻¹
B. Cl⁻² binds 2 electrons, exceeding the octet rule
C. Cl⁺¹, releases 1 electron, remains unstable
D. Cl, the neutral form of Cl, is still unstable with a 7-electron valence configuration
Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂: