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noname [10]
3 years ago
7

What is the volume, in liters, occupied by 0.485 moles of Oxygen gas at 23.0 oC and 0.980 atm?

Chemistry
1 answer:
____ [38]3 years ago
5 0
Use the universal gas formula

PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = 0.08205 L atm / (mol·K)

Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)


Read more on Brainly.com - brainly.com/question/10606064#readmore
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<h3>Explanation</h3>

What's the initial color of the solution?

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\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq).

The first test tube used to contain \text{NH}_4\text{OH}. \text{NH}_4\text{OH} is a weak base that dissociates partially in water.

\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+}  \;(aq)+ {\text{OH}}^{-} \; (aq).

There's also an equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l).

\text{OH}^{-} ions from \text{NH}_4\text{OH} will shift the equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} to the right and reduce the amount of {\text{H}_3\text{O}}^{+} in the solution.

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The equilibrium between  \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions will shift to the left to produce more of both ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)

The indicator equilibrium will shift to the left as the concentration of {\text{H}_3\text{O}}^{+} increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

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3 years ago
Catalyst
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a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

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\tt mass=mol\times MW=1\times 18=18~g

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\tt \dfrac{5}{4}\times 0.34=0.425

Volume O2 at STP :

\tt 0.425\times 22.4=9.52~L

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