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3 years ago

3h2(g)+o3(g) 3h20(g) arrow after the o3(g)whats the overall enthalpy?

Chemistry

1 answer:

Dmitry_Shevchenko [17]3 years ago

8 0

The overall enthalpy of reaction will be calculated from enthalpy of formation of products and reactants.

DeltaHrxn = [Sum of enthalpy of formation of products] - [sum of enthalpy of formation of reactants]

DeltaHrxn = 3 X deltaHf (H2O) - [ 3 X DeltaH (H2) + DeltaH (O3)]

DeltaHrxn = 3 X (--242) - [ 0 + 143] = -869 kJ / mole

We can also calculate the enthalpy of reaction from bond energies

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A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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 $CO_2=13.00g$

Mass of $H_2O=2.662g$

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

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

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 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

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 $CHO_2$

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:

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

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

4 years ago

How many dots should be shown in the Lewis dot diagram of oxygen? (1 point)

deff fn [24]

Answer: the answer is number 3 which is 6

8 0

4 years ago

Find the mass of one atom of uranium-235. Recall that the mass in atomic mass units is equal to the mass in grams of one mole of

stiv31 [10]

Answer:

$3.90*10^{-25}kg/atom$

Explanation:

The molar mass of uranium-235 is 235 g/mol. So one mole of uranium-235 has a mass of 235 g. Put differently 6.022×10^23 atoms of uranium-235 have a mass of 235 g. Knowing that, how can we use that to find the mass of one atom?

mass of one atom = $\frac{235 g}{1mol} *\frac{1 mol}{6.022*10^{23}atoms } *\frac{1kg}{1000g}= 3.90*10^{-25}kg/atom$

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3 years ago

How many moles are in a 62.5-g sample of potassium nitrate (KNO3)?

Reil [10]

Answer:25.3 g of KNO₃ contain 0.25 moles.

Explanation:

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3 years ago

How many moles are in 25.0 grams of KMnO4?

ElenaW [278]

The answer is: 0.158 mol
You find this by doing:
number of moles (n) = mass (m) / molar mass (M)
n=158.034/25.0

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3 years ago