Answer:
81.97 g of NaAl(OH)₄
Solution:
The reaction for the preparation of Sodium Aluminate from Aluminium Metal and NaOH is as follow,
2 NaOH + 2 Al + 6 H₂O → 2 NaAl(OH)₄ + 3 H₂
According to this equation,
2 Moles of NaOH produces = 163.94 g (2 mole) of NaAl(OH)₄
So,
1 Mole of NaOH will produce = X g of NaAl(OH)₄
Solving for X,
X = (1 mol × 163.94 g) ÷ 2 mol
X = 81.97 g of NaAl(OH)₄
_____ are types of active transport.
(A)Diffusion and osmosis
<u>(B)Engulfing and transport proteins
</u>
(C)Osmosis and engulfing
(D)Transport proteins and diffusion
Answer:
C). The structures in the cell work together to perform its life functions.
Explanation:
Unicellular organisms are characterized as organisms that are able to perform all the life functions through a single-cell. The third statement correctly elaborates how '<u>single-celled organisms carry out these life functions by cooperatively employing the structures lying inside the cell</u>.' The first option is incorrect as it talks about multi-cells which doesn't even exist in unicellular organisms. While the second option is wrong because there are no tissues formed of single-cell. The last option is incorrect as the specialized cells perform different life functions in multi-cellular organisms. Thus, <u>option C</u> is the correct answer.
Answer:
La constante de equilibrio Ka del ácido láctico es 1.38x10⁻⁴.
Explanation:
El ácido láctico es un ácido débil cuya reacción de disociación es la siguiente:
CH₃CHOHCOOH + H₂O ⇄ CH₃CHOHCOO⁻ + H₃O⁺ (1)
0.025M - x x x
La constante de acidez del ácido es:
![Ka = \frac{[CH_{3}CHOHCOOH^{-}][H_{3}O^{+}]}{[CH_{3}CHOHCOOH]}](https://tex.z-dn.net/?f=%20Ka%20%3D%20%5Cfrac%7B%5BCH_%7B3%7DCHOHCOOH%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BCH_%7B3%7DCHOHCOOH%5D%7D%20)
Sabemos que la concentración del ácido inicial es:
[CH₃CHOHCOOH] = 0.025 M
Y que a partir del pH podemos hallar [H₃O⁺]:
![pH = -log[H_{3}O^{+}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20)
![[H_{3}O^{+}] = 10^{-pH} = 10^{-2.75} = 1.78 \cdo 10^{-3} M](https://tex.z-dn.net/?f=%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-pH%7D%20%3D%2010%5E%7B-2.75%7D%20%3D%201.78%20%5Ccdo%2010%5E%7B-3%7D%20M%20)
Debido a que el ácido se disocia en agua para producir los iones CH₃CHOHCOO⁻ y H₃O⁺ de igual manera (según la reacción (1)), tenemos:
[CH₃CHOHCOO⁻] = [H₃O⁺] = 1.78x10⁻³ M
Y por esa misma disociación, la concentración del ácido en el equilibrio es:
![[CH_{3}CHOHCOOH^{-}] = 0.025 M - 1.78 \cdo 10^{-3} M = 0.023 M](https://tex.z-dn.net/?f=%20%5BCH_%7B3%7DCHOHCOOH%5E%7B-%7D%5D%20%3D%200.025%20M%20-%201.78%20%5Ccdo%2010%5E%7B-3%7D%20M%20%3D%200.023%20M%20)
Entonces, la constante de equilibrio Ka del ácido láctico es:
Espero que te sea de utilidad!