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2 years ago

Br2(l) + 2Nal(aq) — 12(s) + 2NaBr(aq)

1 answer:

3 0

Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

To determine which elements are oxidized and reduced,

First, we will define the terms Oxidation and Reduction

<em>Oxidation </em>is simply defined as the loss of electrons. It can also be defined as increase in oxidation number.

<em>Reduction </em>is defined as the gain of electrons. It can also be defined as decrease in oxidation number.

The given chemical equation is

Br₂(l) + 2NaI(aq) → I₂(s) + 2NaBr(aq)

Oxidation number of Bromine decreased from 0 to -1.

Therefore, Bromine is reduced.

Oxidation number of Iodine increased from -1 to 0.

Therefore, Iodine is oxidized.

Oxidation number of sodium did not change.

Therefore, Sodium is neither oxidized nor reduced.

Hence, Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.

Learn more here: brainly.com/question/12913997

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How many moles of hydrogen ions are formed in the ionization of 0.250 moles of H2SO4?

True [87]

Answer:

The ionization of 0.250 moles of H₂SO₄ will produce 0.5 moles of H⁺ (hydrogen ion)

Explanation:

From the ionization of H₂SO₄, we have

H₂SO₄ → 2H⁺ + SO₄²⁻

Hence, at 100% yield, one mole of H₂SO₄ produces two moles of H⁺ (hydrogen ion) and one mole of SO₄²⁻ (sulphate ion), therefore, 0.250 moles of H₂SO₄ will produce 2×0.250 moles of H⁺ (hydrogen ion) or 0.5 moles of H⁺ (hydrogen ion) and 0.25 moles of SO₄²⁻ (sulphate ion).

That is; 0.250·H₂SO₄ → 0.5·H⁺ + 0.250·SO₄²⁻.

4 0

4 years ago

A laser produces red light of wavelength 632.8 nm. Calculate the energy,

Ira Lisetskai [31]

Answer:

189.2 KJ

Explanation:

Data Given

wavelength of the light = 632.8 nm

Convert nm to m

1 nm = 1 x 10⁻⁹

632.8 nm = 632.8 x 1 x 10⁻⁹ = 6.328 x 10⁻⁷m

Energy of 1 mole of photon = ?

Solution

Formula used

E = hc/λ

where

E = energy of photon

h = Planck's Constant

Planck's Constant = 6.626 x 10⁻³⁴ Js

c = speed of light

speed of light = 3 × 10⁸ ms⁻¹

λ = wavelength of light

Put values in above equation

E = hc/λ

E = 6.626 x 10⁻³⁴ Js ( 3 × 10⁸ ms⁻¹ / 6.328 x 10⁻⁷m)

E = 6.626 x 10⁻³⁴ Js (4.741 x 10¹⁴s⁻¹)

E = 3.141 x 10⁻¹⁹J

3.141 x 10⁻¹⁹J is energy for one photon

Now we have to find energy of 1 mole of photon

As we know that

1 mole consists of 6.022 x10²³ numbers of photons

So,

Energy for one mole photons = 3.141 x 10⁻¹⁹J x 6.022 x10²³

Energy for one mole photons = 1.89 x 10⁵ J

Now convert J to KJ

1000 J = 1 KJ

1.89 x 10⁵ J = 1.89 x 10⁵ /1000 = 189.2 KJ

So,

energy of one mole of photons = 189.2 KJ

3 0

3 years ago

A 3.82 g piece of limestone contains 2.62 g of CaCO3

Kobotan [32]

Considering the definition of percentage by mass, the mass percentage of CaCO₃ is 68.59%.

<h3>What is mass percentage</h3>

The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

$mass percentage=\frac{mass of solute}{mass of solution}x100$