How many moles in 3.69 x 10^30 molecules of carbon dioxide?

jok3333 [9.3K]

The bond energy of each carbon-oxygen bond in carbon dioxide is d. 736 kJ

Since the chemical reaction is 2CO + O₂ → 2CO₂ and the total bond energy of the products carbon dioxide CO₂ is 1,472 kJ.

Since from the chemical reaction, we have 2 moles of CO₂ which gives 1,472 kJ and there are two carbon-oxygen, C-O bonds in CO₂, then

2 × C-O bond = 1,472 kJ

1 C-O bond = 1.472 kJ/2

C-O bond = 736 kJ

So, the bond energy of each carbon-oxygen bond in carbon dioxide is d. 736 kJ

Learn more about bond energy here:

brainly.com/question/21670527

7 0

2 years ago

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How many atoms of carbon are in the sample

Serhud [2]

There are six atoms in the carbon

6 0

2 years ago

What do the coefficients located before certain molecules in each chemical equation represent?

olga55 [171]

It represents the number of moles required of that molecule to balance the chemical equation, which means to have the reaction chemically happen and goes to completion.

For example:
CH4 + O2 --> H2O + CO2 that is not balanced

with the coefficients located
CH4 + 2O2 --> 2H2O + CO2 now with the coefficients the number of oxygen and hydrogen on each side are equal

6 0

3 years ago

What is the mass of 2.9 moles of calcium? Show all work!

4vir4ik [10]

Answer:

the mass of 2.9 moles of calcium is 116 g

Explanation:

The computation of the mass of 2.9 moles of calcium is shown below

As we know that

Mole = mass ÷ molar mass

where,

Moles be 2.9

And, we know that the molar mass of calcium be 40g/mol

Now put the values to the above formula

2.9 = Mass ÷ 40

So, the mass would be

= 40 × 2.9

= 116 g

Hence, the mass of 2.9 moles of calcium is 116 g

6 0

2 years ago

Never mind jhvjycdtrsesetdfyguhbjnk

OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial pressure of methanol is $P_{CH_3OH_{(g)}} =0.077 \ bar$

The partial pressure of carbon monoxide is $P_{CO} = 0.382 \ bar$

The partial pressure at hydrogen is $P_H = O \ bar$

Explanation:

From the question we are told that

The volume of the flask is $V_f = 10.0 \ L$

The initial pressure of carbon monoxide gas is $P_{CO} = 0.461 \ bar$

The initial temperature of carbon monoxide gas is $T_{CO} = 22.0^oC$

The volume of the hydrogen gas is $V_h = 200 mL = 200 *10^{-3} \ L$

The initial pressure of the hydrogen is $P_H = 7.10 \ bar$

The initial temperature of the hydrogen is $T_H = 271 \ K$

The reaction of carbon monoxide and hydrogen is represented as

$CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}$

Generally from the ideal gas equation the initial number of moles of carbon monoxide is

$n_1 = \frac{P_{CO} * V_f }{RT_{CO}}$

Here R is the gas constant with value $R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K$

=> $n_1 = \frac{0.461 * 10 }{0.0821 * (22 + 273)}$

=> $n_1 = 0.19$

Generally from the ideal gas equation the initial number of moles of Hydrogen is

$n_2 = \frac{P_{H} * V_H }{RT_{H}}$

$n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }$

=> $n_2 = 0.064$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of CO

=> 0.064 moles of hydrogen gas will react with x mole of CO

So

$x = \frac{0.064}{2}$

=> $x = 0.032 \ moles \ of \ CO$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of $CH_3OH_{(g)}$

=> 0.064 moles of hydrogen gas will react with z mole of $CH_3OH_{(g)}$

So

$z = \frac{0.064}{2}$

=> $z = 0.032 \ moles \ of \ CH_3OH_{(g)}$

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is

$n_k = n_1 - x$

=> $n_k = 0.19 - 0.032$

=> $n_k = 0.156$

So at the end of the reaction , the partial pressure for CO is mathematically represented as

$P_{CO} = \frac{n_k * R * T_{CO}}{V}$

=> $P_{CO} = \frac{0.158 * 0.0821 * 295}{10}$

=> $P_{CO} = 0.382 \ bar$

Generally the partial pressure of hydrogen is 0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial pressure of the methanol is mathematically represented as

$P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}$

Here $T_{CO}$ is used because it is given the question that the temperature returned to 22.0 degrees C

So

$P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}$

$P_{CH_3OH_{(g)}} =0.077 \ bar$

6 0

2 years ago

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