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Alchen [17]
4 years ago
12

Write the balanced chemical equation for the reaction shown here. answer chemistry

Chemistry
1 answer:
Aliun [14]4 years ago
7 0
I found the attached image with the same statement of your question and think it may be very useful for you that I use it to show how to answer this question (furthermore I think it may be the same reaction that you forgot to include).

As you can see, there are one image on the left side and other image on hte right side of the figure.

Those images contains drawings that represent molecules and a legend that permit you to distinguish the kind of atoms in each molecule.

Using that, you can indicate the chemical reaction as the transformation of the molecules on the left side onto the molecules on the right side:

Left side:

3 molecules of CH4 and 3 molecules of N2Cl4

Right side:

3 molecules of CCl4, 3 molecules of N2 and 6 molecules of H2

That is represented as:

3CH4 + 3 N2Cl4 -----> 3 CCl4 + 3N2 + 6H2

And that is the balanced chemical equation for the reaction shown in the figured attached.

I hope this is useful for you..






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The presence of 1-2% ethanol as catalyst, suppresses the oxidation of chloroform with oxygen to give a poisonous gas called phosgene. ... Here glycerol acts as negative catalyst. Criteria or characteristics of catalysts. i. The mass and chemical composition of catalyst should remain unchanged at the end of the reaction.

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What is the normal pH range of a Base?
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7.35 - 7.45

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The pH scale ranges from 0 (strongly acidic) to 14 (strongly basic or alkaline). A pH of 7.0, in the middle of this scale, is neutral. Blood is normally slightly basic, with a normal pH range of about 7.35 to 7.45. Usually, the body maintains the pH of blood close to 7.40.

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It is becoming less expensive to screen blood samples for DNA. Certain diseases have a genetic basis. What is a possible ethical
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4 years ago
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Gold has a density of 1,200 lb./ft. What is the density of gold in g/em? For conversion factors use I lb. 453.6 g, and l inch-2.
Jlenok [28]

<u>Answer:</u> The density of gold in g/cm^3 is 19.22g/cm^3

<u>Explanation:</u>

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}

We are given:

Density of gold = 1200lb/ft^3

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into g/cm^3, we get:

\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3

Hence, the density of gold in g/cm^3 is 19.22g/cm^3

6 0
4 years ago
A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
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