In thermodynamics<span>, </span>work<span> performed by a system is the energy transferred by the system to its surroundings. It can be calculated by the expression:
</span>
W = PdV
Integrating,
We will have,
W = P(V2 - V1)
133.7 (1 litre-atm / 101.325 Joule) ( <span>760 Torr / atm ) </span>= 783 (V2 - .0737 )
V2 = 1.35 L
Hope this answers the question. Have a nice day.
Answer:
( °F − 32) × 5/9 = °C
Explanation:
Also there is a mental calculation to convert from Fahrenheit to Celsius. The ratio 5/9 is approximately equal 0.55555….
Subtract 32º to adapt the equivalent in the Fahrenheit scale.
Divide the degrees Celsius by 2 (multiply by 0.5).
Take 1/10 of this number (0.5 * 1/10 = 0.05) and add it to the number obtained previously.
Example: Convert 98.6º F to Centigrade.
98.6 - 32 = 66.6
66.6 * 1/2 = 33.3
33.3 * 1/10 = 3.3
33.3 + 3.3 = 36.6 which is an approximation in degrees Centigrade
To find the concentration of hydronium ions, take 10 raised to the negative pH:
10^-9.56= 2.75 x10^-10M
To find the concentration of hydroxide ions, take 10 raised to the negative pOH: 10^-4.44 = 3.63 x10^-5M
Answer: The value of the equilibrium constant Kc for this reaction is 3.72
Explanation:
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium concentration of 
= 
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For the given chemical reaction:
The expression for is written as:
![K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E3%5Ctimes%20%5BH_2O%5D%5E1%7D%7B%5BHNO_3%5D%5E2%5Ctimes%20%5BNO%5D%5E1%7D)
Thus the value of the equilibrium constant Kc for this reaction is 3.72
