Answer:
1.40 atm
Explanation:
To answer this question we can use<em> Gay-Lussac's law</em>, which states:
When volume and number of moles remain constant.
- T₁ = 23°C ⇒ 23+273.16 = 296.16 K
- T₂ = Boiling point of water = 100 °C ⇒ 100+273.16 = 373.16 K
We <u>put the known data in the equation and solve for P₂</u>:
- 1.11 atm * 373.16 K = P₂ * 296.16 K
What are the answer choices?
Answer:
Moles NH₃: 0.0593
0.104 moles of N₂ remain
Final pressure: 0.163atm
Explanation:
The reaction of nitrogen with hydrogen to produce ammonia is:
N₂ + 3 H₂ → 2 NH₃
Using PV = nRT, moles of N₂ and H₂ are:
N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂
H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂
The complete reaction of N₂ requires:
0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = <em>0.402 moles H₂</em>
That means limiting reactant is H₂. And moles of NH₃ produced are:
0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = <em>0.0593 moles NH₃</em>
Moles of N₂ remain are:
0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = <em>0.104 moles of N₂</em>
And final pressure is:
P = nRT / V
P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L
<em>P = 0.163atm</em>
Explanation:
(a) The molecular equations shows the equation in which all the species of the reactants and the products are in molecules and the net charge is zero.
The complete ionic equations shows the equation in which all the species of the reactants and the products are in dissociated form and are represented as ions.
The net ionic equations shows the equation in which all the species of the reactants and the products are in dissociated form and do not show the spectator ions which are same in the reactants and the products.
(b) If there is no spectator ions in the reaction, then the complete and the net ionic equations would be identical.
Answer:
Two electrons each!
Explanation:
The question pretty much requires us to find the oxidation number of Zinc in the compound.
Zn3P2
Following oxidation number rules;
O.N of Zn3P2 = 0
.Since phosphorus has valence of 5, it needs three more electrons to achieve its octet state. Hence;
Oxidation number of P = -3
Let oxidation number of Zn = x
We have;
3x + 2 (-3) = 0
3x + (-6) = 0
3x = 6
x = 6/3 = 2
This means each zn electrons lost 2 electrons.
