Answer:
54.5 mL
Explanation:
Let's consider the following balanced equation.
Na₂CO₃(aq) + 2 HNO₃(aq) → 2 NaNO₃(aq) + CO₂(g) + H₂O(l)
44.2 mL of 0.108 M Na₂CO₃ solution react. The moles of Na₂CO₃ that react are:
0.0442 L × 0.108 mol/L = 4.77 × 10⁻³ mol
The molar ratio of Na₂CO₃ to HNO₃ is 1:2. The moles of HNO₃ that reacted are 2/1 × 4.77 × 10⁻³ mol = 9.54 × 10⁻³ mol
9.54 × 10⁻³ moles of HNO₃ are in a 0.175 M solution. The required volume of HNO₃ is:
9.54 × 10⁻³ mol × (1 L/0.175 mol) × (1000 mL/1 L) = 54.5 mL
Answer:
Half-life at 310 K is 6.54 × 10³ s.
Explanation:
Let's consider the following reaction:
2 N₂O₅(g) → 4 NO₂(g) + O₂(g)
The rate law is:
(Δ[O₂]/Δt) = k . [N₂O₅]
Since [N₂O₅] is raised to the power of 1, the reaction order is 1.
For a first-order reaction:

where,
is the half-life
is the rate constant
At 300 K,

We can use two-point Arrhenius equation to solve for k₂ at 310 K

At 310 K,

Volume HNO₃ : 51.4 ml
Mass CO₂ : 0.17 g
<h3>Further explanation</h3>
Reaction
Na₂CO₃ + 2 HNO₃ ⇒ 2 NaNO₃ + CO₂ + H₂O
V Na₂CO₃ = 35.7 ml
M Na₂CO₃ = 0.108
mol Na₂CO₃ :

mol ratio Na₂CO₃ : HNO₃ = 1 : 2
mol HNO₃ :

Volume HNO₃ :

mol CO₂ = mol Na₂CO₃ = 3.8556 mlmol=0.00386 mol
mass CO₂(MW=44 g/mol) :

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