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frez [133]
3 years ago
12

why do you think the experimenter conducted this experiment in a lab rather than in the natural world, where this phenomenon was

first observed?
Chemistry
1 answer:
taurus [48]3 years ago
5 0

Answer:

The experimenter observed this experiment in a lab rather than natural world because it might be dangerous to the atmosphere if he does the experiment in the natural world and it was still an hypothesis so that's why he did it in the lab.

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BlackZzzverrR [31]
I’ve only had one and it didn’t flow so ion think so
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7 0
3 years ago
A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o
statuscvo [17]

<u>Answer:</u> The value of K_{eq} is 0.044

<u>Explanation:</u>

We are given:

Initial moles of methane = 4.10\times 10^{-2}mol=0.0410moles

Initial moles of carbon tetrachloride = 6.51\times 10^{-2}mol=0.0651moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of methane = \frac{0.0410}{1.00}=0.0410M

Concentration of carbon tetrachloride = \frac{0.0651}{1.00}=0.0651M

The given chemical equation follows:

                    CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)

<u>Initial:</u>          0.0410    0.0651

<u>At eqllm:</u>     0.0410-x   0.0651-x       2x

We are given:

Equilibrium concentration of carbon tetrachloride = 6.02\times 10^{-2}M=0.0602M

Evaluating the value of 'x', we get:

\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M

Now, equilibrium concentration of methane = 0.0410-x=[0.0410-0.0049]=0.0361M

Equilibrium concentration of CH_2Cl_2=2x=[2\times 0.0049]=0.0098M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}

Putting values in above expression, we get:

K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044

Hence, the value of K_{eq} is 0.044

8 0
3 years ago
Balancee por tanteo las siguientes ecuaciones químicas. Escriba el nombre a reactantes y productos. H2O5 + H2O ---&gt; HNO3 Na2O
Elan Coil [88]

Answer:

a. N₂O₅ + H₂O ⇒ 2 HNO₃ (pentóxido de dinitrógeno + agua ⇒ ácido nítrico)

b. Na₂O + H₂O ⇒ 2 NaOH (óxido de sodio + agua ⇒ hidróxido de sodio)

Explanation:

Tenemos que balancear, por el método de tanteo, las siguientes ecuaciones químicas.

a. En la primera reacción, el pentóxido de dinitrógeno reacciona con agua para formar ácido nítrico. Es una reacción de síntesis o combinación.

N₂O₅ + H₂O ⇒ HNO₃

Podremos obtener la ecuación balanceada si multiplicamos HNO₃ por 2.

N₂O₅ + H₂O ⇒ 2 HNO₃

b. En la segunda reacción, óxido de sodio reacciona con agua para formar hidróxido de sodio. Es una reacción de síntesis o combinación.

Na₂O + H₂O ⇒ NaOH

Podremos obtener la ecuación balanceada si multiplicamos NaOH por 2.

Na₂O + H₂O ⇒ 2 NaOH

3 0
3 years ago
How many moles are in 5.6x1025 molecules of calcium oxide?
Vlad1618 [11]

Answer:

<h2>93.02 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{5.6 \times  {10}^{25} }{6.02  \times  {10}^{23} }  \\  = 93.023...

We have the final answer as

<h3>93.02 moles</h3>

Hope this helps you

8 0
3 years ago
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