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VikaD [51]
3 years ago
4

Liquid benzene burns in the atmosphere. If a cold object is placed directly over the benzene, water will condense on the object

and a deposit of soot (carbon) will also form on the object. The chemical equation for this reaction is of the form x1C6H6 + x2O2 → x3C + x4H2O Determine values of x1, x2, x3, and x4 to balance the equation
Chemistry
1 answer:
Liono4ka [1.6K]3 years ago
4 0

<u>Answer:</u> The value of x_1,x_2,x3\text{ and }x_4 are 2, 3, 12 and 6 respectively.

<u>Explanation:</u>

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.  

A balanced chemical equation always follow this law.

For the given chemical reaction, the balanced equation follows:

2C_6H_6+3O_2\rightarrow 12C+6H_2O

By Stoichiometry of the reaction:

2 moles of benzene reacts with 3 mole of oxygen gas to produce 12 moles of carbon and 6 moles of water

From the above reaction:

x_1=2\\x_2=3\\x_3=12\\x_4=6

Hence, the value of x_1,x_2,x3\text{ and }x_4 are 2, 3, 12 and 6 respectively.

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If you use 1 mole of NaOH, how much NaAl(OH)4 is produced
abruzzese [7]
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
5 0
4 years ago
Need help plzz
Sveta_85 [38]

_____ 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

5 0
3 years ago
What statement best explains how life functions a unicellular organism are carried out?​
lbvjy [14]

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.

8 0
3 years ago
Which of the following statements is true?
kompoz [17]

Answer:

D

Explanation:

7 0
3 years ago
Read 2 more answers
Si tenemos una solución de ácido láctico 0.025M con un pH = 2.75. ¿Cuál es la constante de equilibrio Ka?
rodikova [14]

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]}

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^{+}]

[H_{3}O^{+}] = 10^{-pH} = 10^{-2.75} = 1.78 \cdo 10^{-3} M

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

Entonces, la constante de equilibrio Ka del ácido láctico es:

Ka = \frac{[CH_{3}CHOHCOOH^{-}][H_{3}O^{+}]}{[CH_{3}CHOHCOOH]} = \frac{(1.78 \cdo 10^{-3})^{2}}{0.023} = 1.38 \cdot 10^{-4}          

 

Espero que te sea de utilidad!  

3 0
3 years ago
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