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2 years ago

The substance whose lewis structure shows three covalent bonds is

Chemistry

1 answer:

Gnoma [55]2 years ago

4 0

The substance whose Lewis structure shows three covalent bonds is Nitrogen gas molecule.

<h3>What is Lewis structure?</h3>

Lewis structure is a dot structure which gives idea about the number of valence electrons that are involved in the bonding within the molecule.

Lewis dot structure of nitrogen gas will be expressed as in the attached image, where between two nitrogen atoms triple bond is present. That triple bond is formed by the sharing of electrons and known as covalent bonds.

Hence in nitrogen gas molecule three covalent bonds is present.

To know more about Lewis structure, visit the below link:
brainly.com/question/21026664

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A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt

Oksi-84 [34.3K]

Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

<u>Step 1:</u> Data given

Sample 1: The 1.15 M sample has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1: 1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

<u>Step 3:</u> Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams ( in 55mL)

<u>Step 4</u>: Calculate moles of AgBr

AgNO3 reacts with KBr

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0

3 years ago

You wish to make a 0.285 M hydroiodic acid solution from a stock solution of 12.0 M hydroiodic acid. How much concentrated acid

Inessa05 [86]

The amount, in mL, of the concentrated acid required, would be 1.1875 mL

<h3>Dilution</h3>

From the dilution equation:

m1v1=m2v2 where m1 and m2 = molarity before and after dilution, and v1 and v2 = volume before and after dilution.

m2 = 0.285M, m1 = 12.0M v2 = 50.0 mL

v1 = m2v2/m1 = 0.285x50/12 = 1.1875 mL

Thus, 1.1875 mL of the acid would be taken and diluted with water up to the 50 mL mark.

More on dilution can be found here: brainly.com/question/13949222

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4 0

2 years ago

Confused as heck. please help!

Nitella [24]

Answer: that all thre water cycle and C is vaporation

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bekas [8.4K]

Answer:

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3 years ago

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3 years ago