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3 years ago

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Consider the half reaction below.

1 answer:

Radda [10]3 years ago

7 0

Answer: the second option: <span>Iron is being oxidized
</span>

Explanation:

1) Oxidation is the increase of the oxidation state (number) due to the loss of electrons.

2) In the given reaction, you can see that in the left side the atom is Fe.

When an element (atom) is not combined (or combined with it self) its oxidation state is 0.

3) In the right side of the given equation you that iron is now in form of cation with charge 2+: Fe²⁺.

That means that the new oxidation state of the element is 2+.

4) This change in the oxidation state, of course, is accompanied by the loss of the two electrons: 2e⁻.

5) Conclusion: the iron has oxidized by losing two electrons and increasing its oxidation state from 0 to 2+.
<span></span>

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Vapor obtained by evaporating 0.495 grams of an unknown liquid is collected in a 127 mL flask. At 371 K, the pressure of the vap

lidiya [134]

Answer:

The molar mass in g/mol is 121.4 g/m

Explanation:

Let's apply the Ideal Gases Law to solve this:

P . V = n . R. T

V = 125 mL → 0.125L

P = 754 Torr

760 Torr ___ 1 atm

754 Torr ____ (754 / 760) = 0.992 atm

Moles = Mass / Molar mass

0.992 atm . 0.125L = (0.495 g / MM) . 0.082 . 371K

(0.992 atm . 0.125L) / (0.082 . 371K) = (0.495 g / MM)

4.07x10⁻³ mol = 0.495 g / MM

MM = 0.495 g / 4.07x10⁻³ mol → 121.4 g/m

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How many H2O molecules are in 183.2 grams of H20 gas?

jek_recluse [69]

Answer: There are $61.24 \times 10^{23}$ molecules present in 183.2 grams of $H_{2}O$ gas.

Explanation:

Given: Mass = 183.2 g

Number of moles is the mass of substance divided by its molar mass.

As molar mass of water is 18 g/mol. Therefore, moles of $H_{2}O$ are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{183.2 g}{18 g/mol}\\= 10.17 mol$

According to the mole concept, there are $6.022 \times 10^{22}$ molecules present in one mole of a substance.

Hence, molecules present in 10.17 moles are calculated as follows.

$10.17 mol \times 6.022 \times 10^{23}\\= 61.24 \times 10^{23}$

Thus, we can conclude that there are $61.24 \times 10^{23}$ molecules present in 183.2 grams of $H_{2}O$ gas.

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