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3 years ago

The ionization constant (Ka) of HF is 6.7 X 10 ^-4. Which of the following is true in a 0.1 M solution of this acid?

2 answers:

8 0

The ionization equation is:

HF ⇄ H(+) + F(-)

The ionization constant is Ka = [H(+)] * [H(-)] / [HF]

=> [H(+)] * [F(-)] = Ka * [HF]

Given that Ka < 1

[H(+)] * [F(-)] < [HF]

Which is [HF] > [H(+)] * [F(-)] the option a. fo the list of choices.

n200080 [17]3 years ago

6 0

The correct answer is [HF] is less than [H+][F-]

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Clo3− draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Show t

Karo-lina-s [1.5K]

The formal charges of all nonhydrogen atoms are -1.

Solution:-

<u>O 7-4 = 3 O Double bond on one H 5-4 = 1</u>

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O 3-4= -1 O O 6-7 = -1(2)=-2 H 1-0=+1

It will percentage its last valence electron thru a single bond to the terminal oxygen atom. This is in agreement with carbon and hydrogen atoms that each need to form 4 and 1 covalent bonds respectively. because the terminal oxygen atom best has a single covalent bond, it'll have a proper rate of -1.

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1 year ago

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ZanzabumX [31]

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3 years ago

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What is a type of mechanical weathering in which rocks are broken by the expansion of water as it freezes in joints, pores, or b

Stolb23 [73]

Answer:

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Explanation:

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2 years ago

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A gas has density 2.41 g/liter at 25°C and 770 mm Hg. Calculate it's molecular mass (R = 0.0821 L atm.mol-1K-1

Evgesh-ka [11]

Answer:

Explanation:

Given : Density - 2.41 g/liter

Temperature - 25° C

Pressure : 770 mm Hg

R = 0.0821 L atm mol-¹K-¹

Find : Molecular mass of gas

Solution : Ideal gas equation with respect to density will be : PM = dRT. In the formula, P is pressure, M is molecular mass, d is density, R is gas constant and T is temperature.

Keeping the values in equation-

Pressure : 770 mm Hg = 1 atm

Temperature : 273 + 25 = 298 K

M = dRT/P

M = (2.41*0.0821*298)/1

M = 58.96 gram/mol

Thus, the molecular mass of gas is 58.96 gram/mol.

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3 years ago

How much thermal energy is added to 10.0 g of ice at −20.0°C to convert it to water vapor at 120.0°C?

Sonbull [250]

Answer:

7479 cal.

31262.2 joules

Explanation:

This is a calorimetry problem where water in its three states changes from ice to vapor.

We must use, the calorimetry formula and the formula for latent heat.

Q = m . C . ΔT

Q = Clat . m

First of all, let's determine the heat for ice, before it melts.

10 g . 0.5 cal/g°C ( 0° - (-20°C) = 100 cal

Now, the ice has melted.

Q = Clat heat of fusion . 10 g

Q = 79.7 cal/g . 10 g → 797 cal

We have water at 0°, so this water has to receive heat until it becomes vapor. Let's determine that heat.

Q = m . C . ΔT

Q = 10 g . 1 cal/g°C (100°C - 0°C) → 1000 cal

Water is ready now, to become vapor so let's determine the heat.

Q = Clat heat of vaporization . m

Q = 539.4 cal/g . 10 g → 5394 cal

Finally we have vapor water, so let's determine the heat gained when this vapor changes the T° from 100°C to 120°

Q = m . C . ΔT

Q = 10 g . 0.470 cal/g°C . (120°C - 100°C) → 94 cal

Now, we have to sum all the heat that was added in all the process.

100 cal + 797 cal + 1000 cal + 5394 cal + 94 cal =7479 cal.

We can convert this unit to joules, which is more acceptable for energy terms.

1 cal is 4.18 Joules.

Then, 7479 cal are (7479 . 4.18) = 31262.2 joules

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3 years ago