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3 years ago

In H2SO4, the oxidation number of H is , that of S is , and that of O is .

Chemistry

1 answer:

adell [148]3 years ago

4 0

Answer:

Explanation:

A) In H2SO4, the oxidation number of H is , that of S is , and that of O is .

In H₂SO₄ the oxidation sate of H is +1 (H₂= +2) that of sulfur is +6 and O is -2 (O₄= -8)

In SO₄ the oxidation state is -2, because negative and positive charges are not equal. That's why two hydrogen atoms are attached with one SO₄ to balance the charge.

B) In CaSO4, the oxidation number of Ca is , that of S is . , and that of O is .

In CaSO₄ the oxidation sate of Ca is +2 that of sulfur is +6 and O is -2 (O₄= -8)

In SO₄ the oxidation state is -2, because negative and positive charges are not equal. That's why one calcium atoms is attached with one SO₄ to balance the charge.

C) In HF, the oxidation number of H is , and that of F is .

In HF the oxidation state of hydrogen is +1 and that of fluorine is -1. Both form ionic compound.

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Which equation represents a neutralization reaction?

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Answer: HCI + KOH → KCI + H20

Explanation:

HCI(aq) + KOH(aq) → KCI(aq) + H20(l)

Acid + base → Salt + Water.

The above is a neutralization reaction in which an acid, aqeous HCl reacts completely with an appropriate amount of a base, aqueous KOH to produce salt, aqueous KCl and water, liquid H2O only.

This is a neutralization reaction since, the hydrogen ion, H+, from the HCl is neutralized by the hydroxide ion, OH-, from the KOH to form the water molecule, H2O and salt, KCl only.

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2 years ago

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3 years ago

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Given the following information:

bonufazy [111]

Answer:

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Explanation:

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2 years ago

Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at

valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100 - x

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,

$121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042$

$120.9038x+122.9042\left(100-x\right)=12176.01$

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 % = 42.8 %</u>

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3 years ago

A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme

Simora [160]

Answer:- Density is 10.5 grams per mL and the substance is Silver(Ag).

Solution:- Mass of jewelry is given as 132.6 g. Initial volume of water in the graduated cylinder is 48.6 mL and the volume becomes 61.2 mL when this jewelry is submerged in the cylinder.

So, the volume of the jewelry = 61.2 mL - 48.6mL = 12.6 mL

We know that, $density=\frac{mass}{volume}$

Let's plug in the values in the above formula to calculate the density:

$density=\frac{132.6g}{12.6mL}$

$density=10.5\frac{g}{mL}$

So, the density of the jewelry is $10.5\frac{g}{mL}$ .

The substance from which the jewelry is made could easily be identified by comparing the calculated density with the density values given for different substances in the density table.

Looking at the density table, 10.5 grams per mL is the density for silver. So, the jewelry is made up from Silver.

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3 years ago