All categories

3 years ago

Carbon and oxygen combine to form carbon dioxide. What are the products, or what is the product, in this chemical reaction?

Chemistry

2 answers:

Nata [24]3 years ago

6 0

Answer:

The correct answer is carbon dioxide.

Explanation:

Reagents in a chemical reaction are those that combine to form one or more compounds called products. Carbon and oxygen combine to form carbon dioxide. Experimentally products such as carbon monoxide are also produced, but in the statement, it only indicates that carbon dioxide is produced. For this reason, the product of the reaction is carbon dioxide.

Have a nice day!

VladimirAG [237]3 years ago

4 0

The product in this chemical reaction is Carbon Dioxide or CO2.

This is because it states, "To form," which means that it was produced from the reactants Carbon and Oxygen.

To know what the products are in a statement rather than an equation, words such as produced and formed are used.

Hope this helps!

You might be interested in

Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3

Annette [7]

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving chemical system, in which a reversible chemical reaction takes place. When this situation is reached, it is observed that the concentrations of substances, both reagents and reaction products, they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }$

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

$Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }$

You know:

Kc= 14.5
[H₂]= 0.322 M
[CH₃OH] =1.56 M

Replacing:

$14.5=\frac{1.56}{[CO]*0.322^{2} }$

Solving:

$[CO]=\frac{1.56}{14.5*0.322^{2} }$

[CO]= 1.04 M

The equilibrium value of [CO] is 1.04 M

8 0

3 years ago

Help for this practice work

allochka39001 [22]

Answer:

It's the third option.

Explanation:

In order for the chemical equation to be correctly it needs the same number of atoms of each element on both sides of the equal sign

8 0

2 years ago

What does Le Châtelier's principle state?

bonufazy [111]

When a system experiences a disturbance ( such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state.

5 0

3 years ago

Materials that cannot carry electric energy are called

kompoz [17]

Answer:

They are called insulators

5 0

3 years ago

The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What

pentagon [3]

Answer:

900 J/mol

Explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

= Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C

on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

The heat of vaporization at 75° C = 900 J/mol

8 0

3 years ago