What data should be plotted to show that experimental concentration data fits a secondorder reaction?

A 21.82 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21.33 grams of CO2 and 4

morpeh [17]

Answer: The molecular formula for the given organic compound is $C_2H_2O_4$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=21.33g$

Mass of $H_2O=4.366g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.33 g of carbon dioxide, $\frac{12}{44}\times 21.33=5.82g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.366 g of water, $\frac{2}{18}\times 4.366=0.485g$ of hydrogen will be contained.

Mass of oxygen in the compound = (21.82) - (5.82 + 0.485) = 15.515 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5.82g}{12g/mole}=0.485moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = $\frac{0.485}{0.485}=1$

For Hydrogen = $\frac{0.485}{0.485}=1$

For Oxygen = $\frac{0.969}{0.485}=1.99\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $C_1H_{1}O_2=CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Thus, the molecular formula for the given organic compound is $C_2H_2O_4$

4 0

4 years ago

Which statement is NOT true about Texas during the 1920s? Its C

Hatshy [7]

The answer is c because you said it was c lol

5 0

3 years ago

Suppose that a person eats a diet of 2391 Calories per day.a.) Convert this energy into J. Express your answer using four signif

alekssr [168]

Explanation:

(a)

The given data is:-

Energy = 2391 Calories

The conversion of calories to J is shown below as:-

1 calorie = 4.184 J

So,

Energy = 4.184 * 2391 J = 10003.944 J

Answer in four significant digits: - $1.000\times 10^4\ J$

(b)

The conversion of calories to kJ is shown below as:-

1 calorie = 0.004184 kJ

So,

Energy = 0.004184 * 2391 kJ = 10.003944 kJ

Answer in four significant digits: - $1.000\times 10\ kJ$

(c)

The conversion of calories to kWh is shown below as:-

1 calorie = $1.1622\times 10^{-6}$ kWh

So,

Energy = $1.1622\times 10^{-6}\times 2391$ kWh = 0.002778873 kWh

Answer in four significant digits: - $0.002779\ kWh$

8 0

4 years ago

A reaction vessel contains 10.0 g of CO and 10.0 g of O2. How many grams of CO2 could be produced according to the following rea

iren2701 [21]

Answer:

1. 15.71 g CO2

2. 38.19 % of efficiency

Explanation:

According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:

For CO:

C = 12.01 g/mol

O = 16 g/mol

CO = 28.01 g/mol

(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO

For CO2:

C = 12.01 g/mol

O = 16 x 2 = 32 g/mol

CO2 = 44.01 g/mol

We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:

(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2

We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:

(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2

Now for the efficiency question:

From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,

(6g / 15.71g) x 100 = 38.19 % of efficiency

6 0

3 years ago

What do you think is the effect of acid rain on rusting ?

stich3 [128]

As acid rain is a mixture of pollutants and gasses. As rusting is a chemical reaction between oxygen and other gasses in the air alongside a material and a liquid. I would say they are both similar and that acid rain would speed the process of rusting/aid the process.

8 0

4 years ago