V1/T1 =V2/T2 (using charles law)
V1=6.00
V2=?
T1=273
T2=273
Making V2 the subject of formula the equation then becomes
V2= V1xT2/T1
6.00x263/273=6.0L
Neon is the element that is classified as a noble gas at STP.
Answer:
c. decarboxylation of an a-keto acid.
Explanation:
Decarboxylation refers to the removal of the carboxyl group from a carboxylic acid and thus releasing carbon dioxide. Decarboxylases are enzymes that speed up the removal of the carboxyl group from acids. These reactants could be amino acids, alpha-keto acids, and beta-keto acids. Biotin is known to catalyze the decarboxylation of malonyl CoA to acetyl CoA during fatty acid synthesis.
Malonyl CoA is converted to acetyl CoA after decarboxylation assisted by biotin also known as Vitamin H. Alpha keto acids are involved in fatty acids synthesis and Malonyl CoA is an alpha-keto acid because the keto group is located in the first carbon near the carboxylic acid group. Keto acids have both a carboxyl group and a ketone group.
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
Answer:
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Explanation:
Data Given
M1 = 6.00 M
M2 = 2.5 M
V1 = 250 mL
V2 = ?
Solution:
As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.
Now
first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution
For this Purpose we use the following formula
M1V1=M2V2
Put values from given data in the formula
6 x V1 = 2.5 x 250
Rearrange the equation
V1 = 2.5 x 250 /6
V1 = 104 mL
So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M aqueous solution of ammonium hydroxide
But we have to prepare 250 mL of the solution.
so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.
in this question you have to tell about the amount of water that is 146 mL
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
