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3 years ago

What is the affect of this

Chemistry

2 answers:

maria [59]3 years ago

8 0

Click ok the photo i attatched its the answer

Alisiya [41]3 years ago

5 0

Answer:

Explanation:

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If u were sitting in a boat as 6 meter waves passed by, how many seconds would pass between them

mixer [17]

3374890 seconds would pass

5 0

3 years ago

1. Write a balanced chemical equation for each of the following reactions. Remember that gases of elements such as oxygen are di

Anon25 [30]

Answer:

$CH_4_{(g)}+2O_2_{(g)}\rightarrow CO_2_{(g)}+2H_2O_{(g)}$

$2C_4H_{10}_{(g)}+9O_2_{(g)}\rightarrow 8CO_2_{(g)}+10H_2O_{(g)}$

$H_2SO_4_{(aq)}+2KOH_{(aq)}\rightarrow K_2SO_4_{(aq)}+2H_2O_{(g)}$

Explanation:

(a)

The balanced chemical equation for the reaction of methane with oxygen gas to form carbon dioxide and water is shown below as;

$CH_4_{(g)}+2O_2_{(g)}\rightarrow CO_2_{(g)}+2H_2O_{(g)}$

(b)

The balanced chemical equation for the reaction of butane with oxygen gas to form carbon dioxide and water is shown below as;

$2C_4H_{10}_{(g)}+9O_2_{(g)}\rightarrow 8CO_2_{(g)}+10H_2O_{(g)}$

(c)

The balanced chemical equation for the reaction of solution of sulfuric acid reacts with aqueous potassium hydroxide to produce potassium sulfate and water is shown below as;

$H_2SO_4_{(aq)}+2KOH_{(aq)}\rightarrow K_2SO_4_{(aq)}+2H_2O_{(g)}$

5 0

4 years ago

Determine the oxidation numbers of all the elements in the unbalanced reactions. Then, balance each redox reaction in a basic so

givi [52]

Answer for 1:

- Oxidation number of Mn in MnO4^-1= +7

- Oxidation number of O in MnO4^-1= -2

- Oxidation number of C in C2O4^-2= +3

- Oxidation number of O in C2O4^-2= -2

- Oxidation number of Mn in MnO2= +4

- Oxidation number of O in MnO2= -2

- Oxidation number of C in CO2= +4

- Balanced redox equation in basic solution:

$3C_2O_4^{-2}+2MnO_4^{-1}+4H_2O\operatorname{\rightarrow}6CO_2\text{ +}2MnO_2\text{ + }8OH^-$

Explanation for 1:

1st) To determine the oxidation numbers, it is necessary that the total sum of the charges is equal to that of the molecule or ion in the equation.

• Oxidation numbers in MnO4-,:

We know that the oxidation number for oxygen is -2, then we multiply it by the subscript 4 to find the whole charge of oxygen in this molecule.

Oxygen oxidation number: -2 x 4 = -8

Since the total charge of the molecule is -1, by difference, we will know that the oxidation number of Mn will be +7:

$\begin{gathered} Mn^{+7}O_4^{-2} \\ +7+[(-2*4)]=-1 \end{gathered}$

To confirm that the oxidation number that we determined exists for that element, we can check the Periodic Table of Elements.

We proceed in the same way with all molecules.

• Oxidation numbers in C2O4-2,:

$\begin{gathered} C_2^{+3}O_4^{-2} \\ (+3*2)+[(-2)*4]= \\ +6-8=-2 \end{gathered}$

• Oxidation numbers in MnO2,:

$\begin{gathered} Mn^{+4}O_2^{-2} \\ +4+[(-2)*2]= \\ +4-4=0 \end{gathered}$

• Oxidation numbers in CO2,:

$\begin{gathered} C^{+4}O_2^{-2}_ \\ +4+[(-2)*2]= \\ +4-4=0 \end{gathered}$

2nd) Now that we know the oxidation number os each atom in the reaction, then we can find the element that is oxidized and the element that is reduced.

We can see that Mn goes from +7 to +4, the Mn atom is reduced. And, the carbon atom goes from +3 to +4 so it oxidizes.

3rd) It is necessary to write the oxidation reaction and the reduction reaction separately and balancing all elements except oxygen and hydrogen:

Oxidation:

$C_2O_4^{-2}\text{ }\rightarrow\text{ 2}CO_2\text{ + 2}e^-_$

Reduction:

$MnO_4^{-1}\text{ + 3}e^-\rightarrow MnO_2$

4th) Since the reaction occurs in a basic solution, we must add water (H2O) to balance the oxygen atoms and hydroxyl ion (OH-) to balance the hydrogen atoms. In this case, the reduction reaction is the only one that needs to be balanced with water and hydroxyl ion.

$MnO_4^{-1}\text{ + 3}e^-\text{ +2}H_2O\operatorname{\rightarrow}MnO_2\text{ + 4}OH^-$

5th) It is necessary to balance the electrons in each half-reaction. So, we multiply each half-reaction by the number of electrons in the other half-reaction:

Oxidation:

$\begin{gathered} (C_2O_4^{-2}\operatorname{\rightarrow}\text{2}CO_2\text{ + 2}e_^-)*3 \\ 3C_2O_4^{-2}\operatorname{\rightarrow}6CO_2\text{ + 6}e^- \end{gathered}$

Reduction:

$\begin{gathered} (MnO_4^{-1}\text{ + 3}e^-\text{ + 2}H_2O\operatorname{\rightarrow}\text{ MnO}_2\text{ }+4OH^-)*2 \\ 2MnO_4^{-1}\text{ + 6}e^-\text{ + 4}H_2O\operatorname{\rightarrow}\text{ 2MnO}_2\text{ }+8OH^- \end{gathered}$

6th) We need to cancel out everything that is repeated on opposite sides of the reactions, including the electrons.

Finally, we can write the balanced redox equation:

$3C_2O_4^{-2}+2MnO_4^{-1}+4H_2O\operatorname{\rightarrow}6CO_2\text{ +}2MnO_2\text{ + }8OH^-$

6 0

1 year ago

Ozone is produced by incomplete burning of fuels combustion of sulfur containing fuel decaying organic matter photochemical oxid

VARVARA [1.3K]

Ncomplete combustion of fossil fuels; forest fires// heavy traffic ... NS: oxidation of H2S gasfrom decay of organic matter & volcanic activity ... primary pollutant; burning of sulfur containingfossil fuels, coal containing ... HS: combustion of fossil fuel, industrial plants that produce smoke, ash, dust ..... photochemical smog.

7 0

4 years ago

Read 2 more answers

When 15.0 mL of a 6.42×10-4 M sodium sulfide solution is combined with 25.0 mL of a 2.39×10-4 M manganese(II) acetate solution d

Artist 52 [7]

Answer:

Q = 3.59x10⁻⁸

Yes, precipitate is formed.

Explanation:

The reaction of Na₂S with Mn(CH₃COO)₂ is:

Na₂S(aq) + Mn(CH₃COO)₂(aq) ⇄ MnS(s) + 2 Na(CH₃COO)(aq).

The solubility product of the precipitate produced, MnS, is:

MnS(s) ⇄ Mn²⁺(aq) + S²⁻(aq)

And Ksp is:

Ksp = 1x10⁻¹¹= [Mn²⁺] [S²⁻]

Molar concentration of both ions is:

[Mn²⁺] = 0.015Lₓ (6.42x10⁻⁴mol / L) / (0.015 + 0.025)L = 2.41x10⁻⁴M

[S²⁻] = 0.025Lₓ (2.39x10⁻⁴mol / L) / (0.015 + 0.025)L = 1.49x10⁻⁴M

Reaction quotient under these concentrations is:

Q = [2.41x10⁻⁴M] [1.49x10⁻⁴M]

Q = 3.59x10⁻⁸

As Q > Ksp, the equilibrium will shift to the left producing MnS(s) the precipitate

8 0

4 years ago