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3 years ago

Bromine pentafluoride BrF5 is a good example of inter halogen molecule

Chemistry

1 answer:

Snezhnost [94]3 years ago

3 0

Answer:

a) The lewis dot structure is shown in the image attached to this answer

b) The formal charge on each of the atoms is zero

c) bromine has an oxidation state of +5 while fluorine has an oxidation state of -1

d) 90 degrees

e) Square Pyramidal

f) polar bonds

g) polar molecule

Explanation:

The molecule BrF5 has a formal charge of zero. It exhibits an sp3d2 hybridization state with a square pyramidal geometry. The bond angle in the molecule is 90 degrees. It is a molecule of the type AX5E. The oxidation state of bromine is +5 while that of fluorine is -1.

The Br-F bonds are polar. The overall molecule is polar due to asymmetric charge distribution concentrating on the central atom since the molecule is square pyramidal.

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What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0

expeople1 [14]

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = 45.1mL

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = $\frac{[OH^-][HNO_2]}{[NO_2]}$

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

Where X is reaction coordinate

Replacing in Kb:

1.41x10⁻¹¹ = $\frac{[X][X]}{[0.211 -X]}$

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ FALSE ANSWER. There is no negative concentrations.

X = 1.72x10⁻⁶. Right answer.

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; pH = 8.24

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3 years ago

Hey! I was wondering how to get the solution to this question!

zzz [600]

Answer:

6.22 × 10⁻⁵

Explanation:

Step 1: Write the dissociation reaction

HC₆H₅COO ⇄ C₆H₅COO⁻ + H⁺

Step 2: Calculate the concentration of H⁺

The pH of the solution is 2.78.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.78 = 1.66 × 10⁻³ M

Step 3: Calculate the molar concentration of the benzoic acid

We will use the following expression.

Ca = mass HC₆H₅COO/molar mass HC₆H₅COO × liters of solution

Ca = 0.541 g/(122.12 g/mol) × 0.100 L = 0.0443 M

Step 4: Calculate the acid dissociation constant (Ka) for benzoic acid

We will use the following expression.

Ka = [H⁺]²/Ca

Ka = (1.66 × 10⁻³)²/0.0443 = 6.22 × 10⁻⁵

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3 years ago

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hoa [83]

Answer:

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3 years ago

You need to make 10 mL of 2 mg/ml solution of protein and you have 25 mg/mL solution. How much protein solution and water do you

meriva

Answer:

0.8 mL of protein solution, 9.2 mL of water

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The dilution equation can be used to relate the concentration C₁ and volume V₁ of the stock/undiluted solution to the concentration C₂ and volume V₂ of the diluted solution:

C₁V₁ = C₂V₂

We would like to calculate the value for V₁, the volume of the inital solution that we need to dilute to make the required solution.

V₁ = (C₂V₂) / C₁ = (2mg/mL x 10mL) / (25 mg/mL) = 0.8 mL

Thus, a volume of 0.8 mL of protein solution should be diluted with enough water to bring the total volume to 10 mL. The amount of water needed is:

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3 years ago

11. What is the specific heat of a substance with a mass of 25.5 g that requires 412 J

Romashka-Z-Leto [24]

Answer:

297 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of

1 g

of a given substance by

∘

In your case, aluminium is said to have a specific heat of

0.90

∘

So, what does that tell you?

In order to increase the temperature of

1 g

of aluminium by

∘

, you need to provide it with

0.90 J

of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by

∘

. So if you wanted to increase the temperature of

10.0 g

of aluminium by

∘

, you'd have to provide it with

1 gram



0.90 J

1 gram



0.90 J

...

1 gram



0.90 J



10 times

0.90 J

However, you don't want to increase the temperature of the sample by

∘

, you want to increase it by

∘

−

∘

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

∘



0.90 J

∘



0.90 J

...

∘



0.90 J



33 times

0.90 J

Therefore, the total amount of heat needed to increase the temperature of

10.0 g

of aluminium by

∘

will be

10.0

⋅

0.90

∘

⋅

∘

297 J

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify two sig figs.

For future reference, this equation will come in handy

⋅

, where

- the amount of heat added / removed

- the mass of the substance

- the specific heat of the substance

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

6 0

3 years ago