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galina1969 [7]
3 years ago
10

Consider the following intermediate chemical equations. When you form the final chemical equation, what should you do with CO? C

ancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction. Add the two CO molecules together, and write them as reactants in the final chemical reaction. Write CO only once as a reactant in the final chemical reaction. Write CO as a reactant and a product in the final chemical reaction.
Chemistry
1 answer:
-Dominant- [34]3 years ago
5 0

Answer:

Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.

Explanation:

The CO appears twice hence in he intermediate reaction it only forms path of the enabling reagents and it further reacts to form the final product. Accounting for the CO in the intermediate reaction that undergoes further reaction will impact on the stoichiometry of the reaction.

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Which postulate of Dalton's atomic theory is the result of the law of conservation of mass
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The postulate of Dalton's atomic theory which is a result of the law of conservation of mass is: Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.

8 0
3 years ago
What is the amount of aluminum chloride produced from 40 moles of aluminum and excess chlorine?
Marrrta [24]

Answer:

is it 26.98

Explanation:

6 0
3 years ago
Consider the reaction. PCl5(g)↽−−⇀PCl3(g)+Cl2(g) K=0.042 The concentrations of the products at equilibrium are [PCl3]=0.18 M and
tatyana61 [14]

<u>Answer:</u> The equilibrium concentration of PCl_5 is 1.285 M.

<u>Explanation:</u>

The chemical equation for the decomposition of phosphorus pentachloride follows:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

The expression for equilibrium constant is given as:

K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}

We are given:

K_c=0.042

[PCl_3]=0.18M

[Cl_2]=0.30M

The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Putting values in above equation, we get:

0.042=\frac{0.18\times 0.30}{[PCl_5]}

[PCl_5]=1.285

Hence, the equilibrium concentration of PCl_5 is 1.285 M.

6 0
3 years ago
La suma de los números de masa de tres isótopos es 126 y la suma de los números de neutrones es 60. Hallar la configuración elec
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Ithe correct answer is B good luck
7 0
3 years ago
The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A
s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $ m^3 /d $

                                    $= 80 \times 10^5 \ I/d $

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L $

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d $

If 60 mg of alum $ [Al_2(SO_4)_3.14 H_2O] $ required for one litre of the water treatment.

So Alum required for  $ 80 \times 10^5 \ I/d $

$= 80 \times 15^5 \ I/d  \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d $

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $ = 60 \times 0.234 \text{ of alum ppt. per litre} $

      $= 14.04 $ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

                                        = 112.32 kg/d ppt of alum

Daily total solid load is  $= 144 \ kg/d + 112.32 \ kg/d$

                                       = 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234 $

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

 The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-} $

After addition, the aluminium hydroxide pH of water will increase due to increase in $ OH^- $ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm $

And the pH is reduced into the range of 5.9 to 6.4

6 0
2 years ago
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