Initial Conditions:
Volume= v1= 417 cm³
Temperature= T1 = 278 K
Final Conditions:
Temperature= T2 = 231K
Volume = v2 =?
Use the general gas equation;
P1*v1/T1 = P2*v2/T2
As, the temperature is constant;
So,
v1/T1 = v2/T2
417/278 = v2/231
v2= 346.5 cm³
Answer:
We can't see the options so we don't know what we can put
Explanation:
Answer : The mass of nitric acid is, 214.234 grams.
Solution : Given,
Moles of nitric acid = 3.4 moles
Molar mass of nitric acid = 63.01 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of nitric acid.
Therefore, the mass of nitric acid is, 214.234 grams.
<h3><u>Answer;</u></h3>
<em>-49 °C</em>
<h3><u>Explanation and solution;</u></h3>
- Considering the fact that, the specific heat capacity of aluminum is 0.903 J/g x C, and the heat of vaporization of water at 25 C is 44.0 KJ/mol.
Moles water = 0.48 g / 18.02 g/mol
=0.0266 moles
<em>Heat lost by water</em> = 0.0266 mol x 44.0 kJ/mol
=1.17 kJ => 1170 J
<em>But heat lost =heat gained</em>
<em>Therefore;</em> Heat gained by aluminium = 1170 J
1170 = 55 x 0.903 ( T - 25) = 49.7 T - 1242
1170 + 1242 = 49.7 T
T = 48.5 °C ( 49 °C <em>at two significant figures)</em>
<em>Hence</em>, final temperature = 49 °C
Answer:
The answer to your question is it is not at equilibrium, it will move to the products.
Explanation:
Data
Keq = 2400
Volume = 1 L
moles of NO = 0.024
moles of N₂ = 2
moles of O₂ = 2.6
Process
1.- Determine the concentration of reactants and products
[NO] = 0.024 / 1 = 0.024
[N₂] = 2/1 = 2
[O₂] = 2.6/ 1= 2.6
2.- Balanced chemical reaction
N₂ + O₂ ⇒ 2NO
3.- Write the equation for the equilibrium of this reaction
Keq = [NO]²/[N₂][O₂]
- Substitution
Keq = [0.024]² / [2][2.6]
-Simplification
Keq = 0.000576 / 5.2
-Result
Keq = 1.11 x 10⁻⁴
Conclusion
It is not at equilibrium, it will move to the products because the experimental Keq was lower than the Keq theoretical-
1.11 x 10⁻⁴ < 2400
