Answer:
Explanation:
This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:
The volume of the gas starts at 250 milliliters and the temperature is 137 °C.
The volume of the gas is increased to 425 milliliters, but the temperature is unknown.
We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.
Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.
The units of milliliters (mL) cancel.
The temperature changes to <u>232.9 degrees Celsius.</u>
Answer:
In case of plants surface tension help to support the transpiration pull.
Explanation:
Surface tension is a force that hold the molecules in the surface to minimize the surface area.
During evaporation of excess amount of water from the stomata of leaves of plants by transpiration a surface tension is generated.
The generated surface tension helps to maintain the water column within the xylem tissue by the absorption of water from the soil by the roots.
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![[P(NO_{2} )]^{2}](https://tex.z-dn.net/?f=%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D)
+ 
- 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
Answer:
CH2FCOOH > CH2ClCOOH > CH2BrCOOH > CH3COOH
Explanation:
CH2FCOOH > CH2ClCOOH > CH2BrCOOH > CH3COOH
More electronegative atom of halogen is , stronger acid will be.
