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sesenic [268]
3 years ago
13

Which element is oxidized and which is reduced in the following reactions? (Part A) N2(g) + 3H2(g) →2NH3(g) Express your answers

as chemical symbols separated by a comma. Enter the oxidized element first. (Part B) 3Fe(NO3)2(aq) + 2Al(s) →3Fe(s) + 2Al(NO3)3(aq) Express your answers as chemical symbols separated by a comma. Enter the oxidized element first. (Part C) Cl2(aq) + 2NaI(aq) → I2(aq) + 2NaCl(aq) Express your answers as chemical symbols separated by a comma. Enter the oxidized element first. (Part D) PbS(s) + 4H2O2(aq) →PbSO4(s) + 4H2O(l) Express your answers as chemical symbols separated by a comma. Enter the oxidized element first.
Chemistry
2 answers:
Angelina_Jolie [31]3 years ago
8 0

Answer:

A. H, N

B. Al, Fe

C. I, Cl

D. S, O

Explanation:

To determine if an element is oxidized or reduced we have to consider the change in the oxidation number (ON).

  • If the ON increases, the element is oxidized.
  • If the ON decreases, the element is reduced.

<em>(Part A) </em>

<em>N₂(g) + 3 H₂(g) → 2 NH₃(g) </em>

H is oxidized because its ON increases from 0 to +1.

N is reduced because its ON decreases from 0 to -3.

<em>(Part B) </em>

<em>3 Fe(NO₃)₂(aq) + 2 Al(s) → 3 Fe(s) + 2 Al(NO₃)₃(aq) </em>

Al is oxidized because its ON increases from 0 to +3.

Fe is reduced because its ON decreases from +2 to -0.

<em>(Part C) </em>

<em>Cl₂(aq) + 2 NaI(aq) → I₂(aq) + 2 NaCl(aq) </em>

I is oxidized because its ON increases from -1 to 0.

Cl is reduced because its ON decreases from 0 to -1.

<em>(Part D) </em>

<em>PbS(s) + 4 H₂O₂(aq) → PbSO₄(s) + 4 H₂O(l) </em>

S is oxidized because its ON increases from -2 to +6.

O is reduced because its ON decreases from -1 to -2.

Naya [18.7K]3 years ago
5 0

Answer:

N2(g) + 3H2(g) →2NH3(g)

H, N

3Fe(NO3)2(aq) + 2Al(s) →3Fe(s) + 2Al(NO3)3(aq)

Al, Fe

Cl2(aq) + 2NaI(aq) → I2(aq) + 2NaCl(aq)

I, Cl

PbS(s) + 4H2O2(aq) →PbSO4(s) + 4H2O(l)

S, O

Explanation:

- Determinate the oxidation number in all compounds;  the ox. number that increases means that the element oxidizes, if the number decreases, it is reduced.

N2(g) + 3H2(g) →2NH3(g)

Both elements on the reactives has 0 as ox. number (0 as ground state)

In the ammonia, H acts with +1 and N, with -3

3Fe(NO3)2(aq) + 2Al(s) →3Fe(s) + 2Al(NO3)3(aq)

Fe acts with +2 in reactives, it acts with 0 in products

Al acts with 0 in reactives, it acts with +3 in products

Cl2(aq) + 2NaI(aq) → I2(aq) + 2NaCl(aq)

Cl acts with 0 in reactives, -1 in products

I acts with -1 in reactives, 0 in products

PbS(s) + 4H2O2(aq) →PbSO4(s) + 4H2O(l)

O acts with -1 in reactives, -2 in products

S acts with -2 in reactives, +6 in products

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3 0
3 years ago
classify each of the statements about gases as true or false. oxygen molecules at 25 celsius are moving faster than oxygen
AleksandrR [38]

Since gas molecules average velocity depends on temperature, oxygen molecules at 25°C are moving faster than oxygen molecules at 0 °C.

According to the kinetic theory of gases, the molecules of a gas are in constant random motion and collide frequently with each other and the walls of the container.

The average speed of gas molecules depends on temperature and molar mass as shown by the relation;

vrms = √3RT/M

The following statement are true among the options provided;

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The following are false among the options provided;

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Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days.
8090 [49]

Answer:

11.9g remains after 48.2 days

Explanation:

All isotope decay follows the equation:

ln [A] = -kt + ln [A]₀

<em>Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope</em>

We can find k from half-life as follows:

k = ln 2 / Half-Life

k = ln2 / 27.7 days

k = 0.025 days⁻¹

t = 48.2 days

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ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]

ln[A] = 2.476

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8 0
3 years ago
Limescale removers contain sulfamic acid, H3NSO3, which reacts with minerals in hard water
I am Lyosha [343]

Answer:

0.375 moles of CaCO₃ are required

Explanation:

Given data:

Number of moles of sulfamic acid = 0.75 mol

Number of moles of calcium carbonate required = ?

Solution:

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2H₃NSO₃ + CaCO₃     →        Ca(SO₃NH₂)₂ + CO₂ + H₂O

Now we will compare the moles of H₃NSO₃  and CaCO₃ .

                H₃NSO₃           :            CaCO₃  

                     2                 :             1

                   0.75              :           1/2×0.75 = 0.375 mol

Thus, 0.375 moles of CaCO₃ are required.

 

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