Answer:
Explanation:
1.)azeotrope is a mixture of two or more liquid components under constant boiling, it has a constant mole fraction composition of present component which can be homogeneous or heterogeneous.
2.)the condition which it's best performed when there's liquids that is non-volatile which boils higher than other liquids with at least 26 degrees .
steam azentropic distillation
3.During a steam distillation, How to know if the organic compound is still coming over is when you see the solution becoming cloudy or when there is existence of two layers.
4.)The end of the steam distillation, the receiving flask should contain two layers of liquid, and the chemical identity of these two liquids most contain
A.) Layers that are mostly water H2O
B.) Layers that are mostly products
5.)What is the purpose of adding 10% sodium carbonate solution to the distillate if it is acidic to litmus is to neutralize the distillate.
Answer:
d) Cr⁺³
Explanation:
Consideremos un ion que contiene 24 protones, 28 neutrones y 21 electrones.
Para encontrar el simbolo del elemento, tenemos que considerar el número atómico (Z) que es igual al número de protones. Con esta información, buscamos en la tabla periódica el elemento con Z=24 es el Cromo.
La carga total está dada por la diferencia entre protones y electrones. Los ´protones tienen carga +1 y los electrones carga -1. Luego, este ion tiene carga:
24 protones + 21 electrones = 24 . (1) + 21 . (-1) = +3
El simbolo del ion es Cr⁺³.
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
I think that the answer is 0.49
