Answer:
204.8 K
Explanation:
We use the ideal gas equation:
PV = nRT
where R is the gas constant (0.082 L.atm/K.mol).
We have the following data:
n= 4 moles
P = 5.6 atm
V = 12 L
So, we introduce the data in the ideal gas equation to calculate the temperature (T):
T = PV/nR = (5.6 atm x 12 L)/(4 mol x 0.082 L.atm/K.mol) = 204.8 K ≅ -68 °C
Answer:
Keq: [SO3][NO] /[SO2]NO2]
Explanation:
Complimentary angles are the right answer
Answer:
The true statements are given below.
Explanation:
1 D glucose is a reducing sugar
2 The oxidation of reducing sugar forms a carboxylic acid sugar.
D glucose is a reducing sugar because glucose contain a free hydroxyl group (-OH)in its anomeric carbon.
The oxidation of reducing sugar result in the conversion of -CHO group in case of aldose sugar and -CH2OH group in case of ketose sugar into carboxylic acid(-COOH).
Answer:
pH = 4.09
Explanation:
molarity of oxalic acid in the solution
= 0.1 x 25 / (25 + 35)
= 0.0417 M
molarity of NaOH in the solution
= 0.1 x 35 / (25 +35)
= 0.0583 M
H2C2O4 + NaOH -------------------> NaHC2O4 + H2O
0.0417 0.0583 0 0
0 0.0166 0.0417
now second acid -base titration
NaHC2O4 + NaOH -------------------> Na2C2O4 + H2O
0.0417 0.0166 0 0
0.0251 0 0.0166 ---
now
pH = pKa2 + log [Na2C2O4 / NaHC2O4]
pH = 4.27 + log (0.0166 / 0.0251)
pH = 4.09
