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2 years ago

The Lewis structure for a chlorate ion, ClO3-, should show single bond(s), double bond(s), and lone pair(s).

Chemistry

2 answers:

Yuliya22 [10]2 years ago

7 0

Answer: A molecule of $ClO_3^-$ has 1 single bond, 2 double bonds and 8 lone pairs.

Explanation:

Lewis-dot structure is defined as the structure which represents the number of valence electrons around the atoms. The electrons are represented as dots.

This structure helps in the determination of bonding electrons and non-bonding electrons.

For the given compound $ClO_3^-$

As we know that chlorine has '7' valence electrons and oxygen has '6' valence electron. The negative charge gets add up in the total number of valence electrons.

Total number of valence electrons in $ClO_3^-$ = 7 + 3(6) + 1 = 26

In $ClO_3^-$ , two electrons of 2 oxygen atoms combine with 2 electrons of chlorine and 1 electron of one oxygen atom combines with 1 electron of chlorine atom.

A lone pair remains on the central atom which is chlorine.

Hence, a molecule of $ClO_3^-$ has 1 single bond, 2 double bonds and 8 lone pairs.

eduard2 years ago

3 0

1 single bond, 2 double bond 2 lone pairs

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A 23.0g sample of a compound contains 12.0g of C, 3.0g of H, and 8.0g of O.What the empirical formula of the compound

Kryger [21]

Answer:

The empirical formula of compound is C₂H₆O.

Explanation:

Given data:

Mass of carbon = 12 g

Mass of hydrogen = 3 g

Mass of oxygen = 8 g

Empirical formula of compound = ?

Solution:

First of all we will calculate the gram atom of each elements.

no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms

no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms

no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms

Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.

C:H:O = 1/0.5 : 3/0.5 : 0.5/0.5

C:H:O = 2 : 6 : 1

The empirical formula of compound will be C₂H₆O

5 0

2 years ago

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c

Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

$18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M$

Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

$42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M$

6 0

2 years ago

A solution is made by dissolving 58.125 g of sample of an unknown, nonelectrolyte compound in water. The mass of the solution is

e-lub [12.9K]

Answer:

molecular weight (Mb) = 0.42 g/mol

Explanation:

mass sample (solute) (wb) = 58.125 g

mass sln = 750.0 g = mass solute + mass solvent

∴ solute (b) unknown nonelectrolyte compound

∴ solvent (a): water

⇒ mb = mol solute/Kg solvent (nb/wa)

boiling point:

ΔT = K*mb = 100.220°C ≅ 373.22 K

∴ K water = 1.86 K.Kg/mol

⇒ Mb = ? (molecular weight) (wb/nb)

⇒ mb = ΔT / K

⇒ mb = (373.22 K) / (1.86 K.Kg/mol)

⇒ mb = 200.656 mol/Kg

∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg

moles solute:

⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute

molecular weight:

⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol

8 0

2 years ago

Give the product for the reaction of 1-butene with methanol in the presence of acid. The mechanism is the same as the mechanism

Nuetrik [128]

Answer:

The final result is 2-methoxybutane.

Explanation:

1-butene has a carbon-carbon double bound between C1 and C2.

When it will react with methanol, in the presence of an acid, the result will be an ether.

C4H8 + CH3OH → C5H12O

An acid-catalyzed ether synthesis from alkenes is limited by carbocation stability.

In the first step, the double bound will disappear. The C2 atom will be a C+ atom, this becaus it has only 3 bounds and not 4.

This C+ -atom will atract the O- atom to form an ether. The CH3 of methanol will bind on the C3 atom, this is the most stable position.

2-methoxybutane will be formed. It has a structural formula of C5H12O

1-methoxybutane will not be formed, because it's less stable.

1-butanol will be formed when water is added to 1-butene. The mechanism has the same principle but not the same product.

1-ethoxybutane and 2-ethoxybutane have a structural formula of C6H14O, this will not be the final result.

The final result is 2-methoxybutane.

8 0

2 years ago

Calculate the mass in grams of each of the following amounts: 1.002 mol of chromium 4.08 x 10-8 mol of neon

Pepsi [2]

Answer:

$Mass_{chromium}=52.1\ g$

$Mass_{neon}=8.23\times 10^{-7}\ g$

Explanation:

Calculation of the mass of chromium as:-

Moles = 1.002 moles

Molar mass of chromium = 51.9961 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$1.002\ mol= \frac{Mass}{51.9961\ g/mol}$

$Mass_{chromium}=1.002\times 51.9961\ g = 52.1\ g$

Calculation of the mass of neon as:-

Moles = $4.08\times 10^{-8}$ moles

Molar mass of neon = 20.1797 g/mol

Thus,

$1.002\ mol= \frac{Mass}{20.1797\ g/mol}$

$Mass_{neon}=4.08\times 10^{-8}\times 20.1797\ g = 8.23\times 10^{-7}\ g$

6 0

2 years ago

The Lewis structure for a chlorate ion, ClO3-, should show ____ single bond(s), ____ double bond(s), and ____ lone pair(s).

The Lewis structure for a chlorate ion, ClO3-, should show single bond(s), double bond(s), and lone pair(s).