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3 years ago

In the chemical reaction shown below how do you classify hydrogen and oxygen 2h2+o2→h2o

Chemistry

1 answer:

earnstyle [38]3 years ago

6 0

In the chemical reaction shown, 2H2 + O2→H2O, hydrogen and oxygen are called reactants.

A chemical reaction equation is has a left and a right hand side separated by an arrow.

The species on the left hand side "combine" to yield the specie(s) on the right hand side.

These species on the left hand side undergo breaking of bonds are rearrangement of atoms to give the species on the right hand side.

Hence, the species on the left hand side are called reactants because they combine with each other to give the species on the right hand side (products).

Learn more: brainly.com/question/17096236

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The combustion of a sample of butune, C4H10 (lighter fluid), produced 2.46 grams of water.

Marizza181 [45]

Molar mass :

H₂O = 18.0 g/mol
O₂ = 32.0 g/mol

2 C₄H₁₀ + 13 O₂ = 8 CO + 10 H₂O

13 x 32.0 g O₂ ------------- 10 x 18.0 g H₂O
? g O₂ ------------- 2.46 g H₂O

Mass O₂ = 2.46 x 13 x 32.0 / 10 x 18.0

Mass O₂ =1023.36 / 180

=> 5.68 g of O₂

hope this helps!

7 0

3 years ago

Consider the reaction: CO2(g) + CCl4(g) ⇌ 2 COCl2(g) ΔG° = 46.9 kJ Under the following conditions at 25 oC: LaTeX: P_{CO_2}P C O

Mamont248 [21]

Answer : The value of $\Delta G_{rxn}$ is, $-47.0kJ/mole$

Explanation :

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln K_p$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction

$\Delta G_^o$ = standard Gibbs free energy = 46.9 kJ

R = gas constant = 8.314 J/mole.K

T = temperature = $25^oC=273+25=298K$

$K_p$ = equilibrium constant

First we have to calculate the value of $K_p$ .

The given balanced chemical reaction is,

$CO_2(g)+CCl_4(g)\rightarrow 2COCl_2(g)$

The expression for reaction quotient will be :

$K_p=\frac{(p_{COCl_2})^2}{(p_{CO_2})\times (p_{CCl_4})}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$K_p=\frac{(0.653)^2}{(0.459)\times (0.984)}$

$K_p=0.944$

Now we have to calculate the value of $\Delta G_{rxn}$ by using relation (1).

$\Delta G_{rxn}=\Delta G^o+RT\ln K_p$

Now put all the given values in this formula, we get:

$\Delta G_{rxn}=-46.9kJ/mol+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (0.944)$

$\Delta G_{rxn}=-47.0kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is, $-47.0kJ/mole$

3 0

3 years ago

Explain how an increase in the temperature causes an increase in the solubility of most solids and liquids

Talja [164]

Increasing temperature can cause the molecules in liquid to spread further apart allowing the mixture to be more soluble.

7 0

4 years ago

A carbocation is an example of a(an) ____. Does a carbocation accept a pair of electrons, or does it donate a pair of electrons?

fgiga [73]

Answer:

Option A. The carbocation accepts a pair of electrons.

Explanation:

A carbocation is defined as a positively charged carbon, which is bound to 3 substituents. Since it has no electrons nonbonding, it only has six electrons in its valence shell. With only six electrons in its valence shell, a carbocation is a powerful electrophile (and Lewis acid) and can react with any nucleophile that is found.

Carbocations are proposed as intermediates in many organic reactions. They also work like free radicals, which are electron-deficient species.

Same as free radicals, the carbocations are stabilized by alkyl substituents.

8 0

3 years ago

What happens to the energy released by an exothermic reaction

Stels [109]

Answer: 1

Explanation:

5 0

4 years ago

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