Answer:
structure of argon
Explanation:
The nucleus contains 18 protons ( red) and 22 neutrons (orange). 18 electrons (white) occupy shells ring.
Answer:
P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)
Explanation:
Given:
P₁ = 90 atm P₂ = ?
V₁ = 18 Liters(L) L₂ = 12 Liters(L)
=> decrease volume => increase pressure
=> volume ratio that will increase 90 atm is (18L/12L)
T₁ = 272 Kelvin(K) T₂ = 274 Kelvin(K)
=> increase temperature => increase pressure
=> temperature ratio that will increase 90 atm is (274K/272K)
n₁ = moles = constant n₂ = n₁ = constant
P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)
By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.
The new volume when pressure increases to 2,030 kPa is 0.8L
BOYLE'S LAW:
The new volume of a gas can be calculated using Boyle's law equation:
P1V1 = P2V2
Where;
- P1 = initial pressure (kPa)
- P2 = final pressure (kPa)
- V1 = initial volume (L)
- V2 = final volume (L)
According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:
406 × 4 = 2030 × V2
1624 = 2030V2
V2 = 1624 ÷ 2030
V2 = 0.8L
Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.
Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!
