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3 years ago

Which carbon atom has the most partial positive character?

1 answer:

4 0

A carbon atom would have a partial positive character if it is bonded with a much more electronegative element. The periodic trend for electronegativity is increasing as you go from right to left, and decreasing as you go down a group. Thus, a more electronegative element than carbon would either be nitrogen, oxygen and fluorine. Hence, bonds involving CN, CO and CF would have a partial positive character for carbon.

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Sodium peroxide (Na2O2) is often used in self-contained breathing devices, such as those used in fire emergencies, because it re

Anastaziya [24]

Answer:

725.15 L

Explanation:

The balanced chemical equation for the reaction between Na₂O₂ and CO₂ is the following:

Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂

From the stoichiometry of the reaction, 1 mol of Na₂O₂ reacts with 1 mol CO₂. So, the stoichiometric ratio is 1 mol Na₂O₂/1 CO₂.

Now, we convert the mass of reactants to moles by using the molecular weight (Mw) of each compound:

Mw (Na₂O₂) = (23 g/mol x 2) + (16 g/mol x 2) = 78 g/mol

moles Na₂O₂ = 96.7 g/(78 g/mol) = 1.24 mol Na₂O₂

Mw (CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol

moles CO₂ = 0.0755 g/(44 g/mol) = 1.71 x 10⁻³ mol CO₂

Now, we calculate the number of moles of CO₂ we need to completely react with the mass of Na₂O₂ we have:

1.24 mol Na₂O₂ x 1 mol CO₂/1 mol Na₂O₂ = 1.24 mol CO₂

In 1 L of respired air we have 1.71 x 10⁻³ mol CO₂ (0.0755 g), so we need the following number of liters to have 1.24 mol of CO₂:

1.24 mol CO₂ x 1 L/1.71 x 10⁻³ mol CO₂ = 725.15 L

4 0

3 years ago

How much of sliver metal will weight exactly 2500.0

grandymaker [24]

The density of silver is 10.5 g/cm3.

8 0

3 years ago

Copper can be broken into simpler substances so what is it

allochka39001 [22]

Cooper cannot be broken down into a simpler substance since it is an element meaning it had its own atoms.

6 0

3 years ago

State Hess' law of constant heat summation.

AveGali [126]

Answer:

-74.6 kj/mol

Explanation:

you can see the answer at the pic

8 0

3 years ago

Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

Molarity of CH4:

CH4 = $\frac{20}{125}$ = 0.16M

Molarity of H20:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

Molarity of H2:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

Molarity of CO:

CO = $\frac{0.144}{3}$ = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

Molarity of CH4:

CH4 = 0.16 - 0.048 = 0.112M

Molarity of H2O:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

$K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }$

$K_{c}$ = $\frac{0.048.0.144^{3} }{0.112.0.032}$

$K_{c}$ = 2. $10^{-2}$

The concentration equilibrium constant for the process is $K_{c}$ = 2. $10^{-2}$ .

4 0

3 years ago