Sort the descriptions below to indicate whether they describe β oxidation of stearoyl-coa, oleoyl-coa, or both. items (6 items)
1 answer:
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Charles law states that volume of gas is directly proportional to temperature at constant pressure.
where V - volume and T - temperature
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - 32 °C + 273 = 305 K
substituting these values in the equation
T = 149 K
temperature in celsius = 149 K - 273 = - 124 °C
new temperature is 124 °C
where V - volume and T - temperature
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - 32 °C + 273 = 305 K
substituting these values in the equation
T = 149 K
temperature in celsius = 149 K - 273 = - 124 °C
new temperature is 124 °C
Answer:
The temperature doesn't change
Explanation:
Given that:
Where
- U: internal energy
- W: work
- Q: heat
If the gas does no work and doesn't exchange heat with the outside there insn't a variation of the internal energy.
Internal energy is realted to the energy in the molecules, they vibration and theregore the temperature.
So, if there isn't a change in the internal energy you may say that there isn't a change in the temperature.
The increase in the volume is balanced by a decrease of the pression.
Answer:
1) The dilution scheme will result in a 200μM solution.
2) The dilution scheme will not result in a 200μM solution.
3) The dilution scheme will not result in a 200μM solution.
4) The dilution scheme will result in a 200μM solution.
5) The dilution scheme will result in a 200μM solution.
Explanation:
Convert the given original molarity to molar as follows.
Consider the following serial dilutions.
1)
Dilute 5.00 mL of the stock solution upto 500 mL . Then dilute 10.00 mL of the resulting solution upto 250.0 mL.
<u>Molarity of 500 mL solution:</u>
<u>10 mL of this solution is diluted to 250 ml</u>
<u>Convert μM</u> :
Therefore, The dilution scheme will result in a 200μM solution.
2)
Dilute 5.00 mL of the stock solution upto 100 mL . Then dilute 10.00 mL of the resulting solution upto 1000 mL.
<u>Molarity of 100 mL solution:</u>
<u>10 mL of this solution is diluted to 1000 ml</u>
<u>Convert μM</u> :
Therefore, The dilution scheme will not result in a 200μM solution.
3)
Dilute 10.00 mL of the stock solution upto 100 mL . Then dilute 5 mL of the resulting solution upto 100 mL.
<u>Molarity of 100 mL solution:</u>
<u>5 mL of this solution is diluted to 1000 ml</u>
<u>Convert μM</u> :
Therefore, The dilution scheme will not result in a 200μM solution.
4)
Dilute 5 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 500 mL.
<u>Molarity of 250 mL solution:</u>
<u>10 mL of this solution is diluted to 500 ml</u>
<u>Convert μM</u> :
Therefore, The dilution scheme will result in a 200μM solution.
5)
Dilute 10 mL of the stock solution upto 250 mL . Then dilute 10 mL of the resulting solution upto 1000 mL.
<u>Molarity of 250 mL solution:</u>
<u>10 mL of this solution is diluted to 1000 ml</u>
<u>Convert μM</u> :
Therefore, The dilution scheme will result in a 200μM solution.