Answer:
NO is the limiting reagent.
In this reaction 0.886 mole of NO2 is produced
Explanation:
The chemical equation for this reaction is
2NO(g) + O2(g) → 2NO2(g)
In this limiting reagent reaction, 2 moles of NO reacts with one mole of O2 to produce 2 mole of 2NO2
0.886 mole of NO * (2 mole of NO2/2 mole of NO) = 0.886 mole of NO2
0.503 mole of O2 * (1 mole of NO2/1 mole of O2) = 1.01 mole of NO2
Hence, NO is the limiting reagent.
In this reaction 0.886 mole of NO2 is produced
Volume of NaOH required to react = 145.5 ml
<h3>Further explanation</h3>
Reaction
CO₂(g)
+ 2 NaOH(aq) ⇒Na₂CO₃(aq) + H₂O(l)
The volume of CO₂ : 0.45 L
mol CO₂ at STP (O C, 1 atm) ⇒ at STP 1 mol gas 22.4 L :
From the equation, the mol ratio of CO₂ : NaOH = 1 : 2, so mol NaOH :
Then volume of NaOH :
In chemistry, neutralization or neutralisation (see spelling differences) is a chemical reaction in which an acid and a base react quantitatively with each other. In a reaction in water, neutralization results in there being no excess of hydrogen or hydroxide ions present in the solution.
Given:3.40g sample of the steel used to produce 250.0 mLSolution containing Cr2O72−
Assuming all the Cr is contained in the BaCrO4 at the end.
(0.145 g BaCrO4) / (253.3216 g BaCrO4/mol) x (250.0 mL / 10.0 mL) x (1 mol Cr / 1 mol BaCrO4) x (51.99616 g Cr/mol / (3.40 g) = 0.219 = 21.9% Cr
The energy range expected is 6.6 × 10^-19 J < E < 7.33 × 10^-19 J
The energy of the photon is given by;
E = hc/λ
E = energy of the photon
h = Plank's constant
c = speed of light
λ = wavelength of light
For the upper boundary range;
E = ?
h = 6.6 × 10^-34 Js
c = 3 × 10^8 m/s
λ = 270 × 10^-9
E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 270 × 10^-9
E = 7.33 × 10^-19 J
For the lower range;
E = ?
h = 6.6 × 10^-34 Js
c = 3 × 10^8 m/s
λ =300 × 10^-9
E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 300 × 10^-9
E = 6.6 × 10^-19 J
Hence, the energy range 6.6 × 10^-19 J < E < 7.33 × 10^-19 J
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