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Dafna1 [17]
3 years ago
9

Which balanced equation represents an oxidation-reduction reaction

Chemistry
1 answer:
geniusboy [140]3 years ago
6 0

The question is incomplete, the complete question is;

Which balanced equation represents an oxidation-reduction reaction

A) AgNO3(aq) + NaCI(aq) ------>AgCI(s) + NaNO3(aq)

B) H2CO3(aq) -------> H2O(l) + CO2(g)

C) NaOH(aq) + HCl(aq) ----------> NaCl(aq) + H2O(l)

D) Mg(s) + 2HCl(aq) -------->MgCl2(aq) + H2(g)

Answer:

D) Mg(s) + 2HCl(aq) -------->MgCl2(aq) + H2(g)

Explanation:

Redox reaction is a type of chemical reaction in chemistry in which the oxidation states of atoms are changed. Redox reactions are characterized by an actual transfer of electrons between chemical species. An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron. Thus Oxidation-reduction reaction, refers to any chemical reaction in which the oxidation number of a participating chemical species changes. The term covers a large and diverse body of processes.

If we look at option D very closely, the reaction is written as;

Mg(s) + 2HCl(aq) -------->MgCl2(aq) + H2(g)

We can clearly see that from left to right;

The oxidation number of magnesium increased from zero to +2

Oxidation number of hydrogen decreased from +1 to zero.

Hence option D depicts an oxidation-reduction reaction.

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We are given the number of moles of solid magnesium supplied for the reaction which is 0.02 moles while hydrochloric acid is supplied in excess thus we can say that the reaction proceeds to completion. Calculation is as follows:

0.020 mol Mg ( 1 mol H2 / 1 mol Mg ) = 0.020 mol H2 gas is produced

To convert the number of moles to volume, we use the conditions at STP of 1 mol of a substance is equal to 22.4 L. Thus,

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calculate the amount of time required to rise the temperature of 0.50 liter of water from 0.0 c to 10.0 c if the energy is suppl
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Answer:

The amount of time required to rise the temperature of 0.50 liter of water from 0.0 c to 10.0 c is 19 seconds

Explanation:

1100 J/s is the energy supplied so, we have to set, the energy to that ΔT, which is 10°C (10°-0°)

Q = m . C. ΔT

Q =  500g . 4,186J/g°C . 10°C

Remember the water density is 1g/mL, so if we have 0,5 L, we should get the convertion to g. (1ml = 1x10*-3 L)

1x10*-3 L ____ 1 g

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4,186J/g°C, this value is known as specific heat of water. It is a known value.

Q =  500g . 4,186J/g°C . 10°C = 20930 J

Now the final rule of three:

1100 J supplies at _____ 1 second

20930 J supplies at _____ (20930 J . 1 s) / 1100 J =19,02 s

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