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Alex
2 years ago
15

When carbon-containing compounds are burned in a limited amount of air, some CO(g) as well as CO₂(g) is produced. A gaseous prod

uct mixture is 35.0 mass % CO and 65.0 mass % CO₂. What is the mass % of C in the mixture?
Chemistry
1 answer:
solong [7]2 years ago
3 0

Here, we are going to calculate the mass % of C in the mixture.

What is a Mixture?

A mixture is composed of one or more pure substances in varying composition. There are two types of mixtures: heterogeneous and homogeneous. Heterogeneous mixtures have visually distinguishable components, while homogeneous mixtures appear uniform throughout.

Given that,

The mass % of CO =35.0% =35.0 g in 100 g mixture

The mass % of CO2 = 65% =65 g in 100 g mixture

Therefore,

The mass of C from CO = 15.007 g C

Similarly,

The mass of C from CO2 = 17.738 g C

Thus, the total mass of C = 15.007 g+17.738 g =32.745 g

Therefore,

The mass % of C= 32.745% =32.7%

Thus, the mass % of C in the mixture is 32.7%

To learn more about carbon-containing compounds click on the link below:

brainly.com/question/13381262

#SPJ4

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If Liquid 1 has a higher specific heat than Liquid 2, then Liquid 1 will take longer to increase in temperature because the higher specific heat of a liquid needs more thermal energy for heating a liquid.

<h3>What is specific heat?</h3>

Specific heat of a substance refers to the quantity of heat that is required to raise the temperature of one gram of a substance by one Celsius degree so we can conclude that  Liquid 1 will take longer to increase in temperature

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3 years ago
How do the lithosphere and the asthenosphere work together?
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2 years ago
In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b
kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

Explanation:

Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)

The reaction is first order in [Co(NH_3)5Br]^{2+}:

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=0.100 M

a) Final concentration of [Co(NH_3)5Br]^{2+} after 19.0 hours= [A]

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}

[A]=0.065 M

0.065 M is its molarity after a reaction time of 19.0 h.

b)

Initial concentration of [Co(NH_3)5Br]^{2+}= [A_o]=x

Final concentration of [Co(NH_3)5Br]^{2+} after t = [A]=(100\%-69\%) x=31\%x=0.31x

Rate constant of the reaction = k = 6.3\times 10^{-6} s^{-1}

The integrated law of first order kinetic is given as:

[A]=[A_o]\times e^{-kt}

0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}

t = 185,902.06 s = \frac{185,902.06 }{3600} hour = 51.64 hours ≈ 52 hours

In 52 hours [Co(NH_3)5Br]^{2+} will react 69% of its initial concentration.

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Consider the redox reaction below.
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Answer:

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Explanation:

Let us consider the complete redox reaction:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

  • Oxidation (loss of electrons or increase in the oxidation state of entity)
  • Reduction (gain of electrons or decrease in the oxidation state of the entity)
  • An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.

Here Zn(s) is undergoing oxidation from OS 0 to +2

And H in HCl (aq) is undergoing reduction from OS +1 to 0.

Therefore, for this reaction;

Oxidation Half equation is:

Zn(s) → Zn⁺²(aq) + 2e⁻

Reduction Half equation is:

2H⁺ + 2e⁻ → H₂(g)

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