Answer:
c.corn syrup
the egg shrinks due to osmosi.
The final temperature : 345 K
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Further explanation
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Given
475 cm³ initial volume
600 cm³ final volume
Required
The final temperature
Solution
At standard temperature and pressure , T = 273 K and 1 atm
Charles's Law :
When the gas pressure is kept constant, the gas volume is proportional to the temperature
V₁/T₁=V₂/T₂
Input the value :
T₂=(V₂T₁)/V₁
T₂=(600 x 273)/475
T₂=345 K
Answer:
C
Explanation:
A fuel cell uses both hydrogen and oxygen as fuels. At the cathode oxygen is reduced to water according to the equation
O2+4H^+ +4e ------>2H2O
And hydrogen is oxidized at the anode:
2H2------>4H^+ +4e
Fuel cells produce only water as a by product hence they are environment friendly
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp = 
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104
+
- 200 = 0
Resolving the second degree equation:
=
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are
= 98.7 MPa and P(N₂O₄) = 101.3 MPa