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3 years ago

The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?

Chemistry

2 answers:

Mazyrski [523]3 years ago

8 0

Last option that is none of above is right answer.

MrRa [10]3 years ago

7 0

Answer : The correct option is, $3.5\times 10^{-4}$

Solution : Given,

pH = 2.03

Concentration of HF = 0.25 M

First we have to calculate the concentration of $H^+$ ion.

$pH=-\log [H^+]$

$2.03=-\log [H^+]$

$[H^+]=9.3\times 10^{-3}M$

Now we have to calculate the value of $K_a$ for HF.

The equilibrium reaction will be

$HF\rightleftharpoons H^++F^-$

Concentration of $H^+$ = Concentration of $F^-$ = $9.3\times 10^{-3}M$

The expression for $K_a$ for HF will be,

$K_a=\frac{[H^+][F^-]}{[HF]}=\frac{(9.3\times 10^{-3})\times (9.3\times 10^{-3})}{0.25}=3.5\times 10^{-4}$

Therefore, the value of $K_a$ for HF is, $3.5\times 10^{-4}$

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whick of the following states of matter is characterized by having definite shape a)liquid b)gas c)solid d)plasma

zavuch27 [327]

Answer:

c) A solid has a definite shape

Explanation:

Only solids have definite shape. Liquids acquire the shape of the container and gases have molecules that move randomly .Gases turn into plasma when

heat energy is added to it.

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3 years ago

Read 2 more answers

Given 4.80g of ammonium carbonate, find:

V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

1) Number of moles of the compound:

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

2) Number of moles of ammonium ions :

Ammonium carbonate is dissociated according to the balanced equation:

(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ = (2.0)(0.05 mol) = 0.1 mol.

3) Number of moles of carbonate ions :

Ammonium carbonate is dissociated according to the balanced equation:

(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

4) Number of moles of hydrogen atoms:

Every 1.0 mol of (NH₄)₂CO₃ contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO₃ = (8.0)(0.05 mol) = 0.4 mol.

5) Number of hydrogen atoms:

It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

Using cross multiplication:

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

∴ The no. of atoms in 0.4 mol of H atoms = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = 2.4 x 10²³ molecules.

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3 years ago

1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change

Inga [223]

Answer:

1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

Explanation:

1) To show that heat flows spontaneously from high temperature to low temperature

example :

Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )

Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment

Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )

Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )

Δs ( entropy change for cold block ) = Q / tc

∴ Total Δs = ΔSc + ΔSh

= Q/tc - Q/th

2) Entropy change for Decomposition of mercuric oxide

2HgO (s) → 2Hg(l) + O₂ (g)

Δs = positive

there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C

hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =

Δh of reaction = 181.6 KJ

Temp = 500 + 273 = 773 k

hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k

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3 years ago

___________________is the rate that an object moves in a certain direction. Question 3 options:

Nesterboy [21]

Answer: Velocity

Explanation:

i just took the quiz for k12

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3 years ago

Instructions

ivann1987 [24]

Answer:

I got a 100 with this, sorry if this is not what you want just trying to help

Explanation:

1. This experiment was to find how mass and speed effect KE. This is important because if you were in a situation where you needed something to go higher, you would know to add more or less of mass/speed.

To test mass, we filled the bean bag with a certain amount of water, then dropped it. After, you recorded how high it made the bean bag go. The same with speed, but same amount in the bottle, just dropped from different heights.

My hypothesis is when you have more mass, the KE will be greater. This is also the same with speed, if it is dropped from a higher place, the bean bag will launch farther than the last time.

2. Data I collected from the lab was like my hypothesis explained. When the height of the bottle increased, it made the bean bag go higher than the last. And I tested 4 different masses, 0.125 kg, 0.250kg, 0.375kg and 0.500kg. Each time the bean bag went higher on a larger mass.

A lot of times on the speed test, the bean bag would go higher than the bottle drop point, but not every time. Also, when it was dropped from the same height each time, some results varied quite a bit, like when it was dropped from 1.28 the results were 1.14 then 1.30 1.30. Mass on the other hand was all in the same number range, only once the numbers were a bit off from each other.

3. Some formulas I used were KE= ½ mv^2 and Ht v^2/2g. The first was to calculate the kinetic energy of an object, m=mass v=speed. Second was for finding out what height I needed to drop something to reach a certain speed, Ht=Height and g= Gravitational Acceleration of 9.8 m/s^2.

I used these to figure out tables that showed relationships between different things like mass and KE or speed and height. The whole time I was doing the lab, my data was going up, when there was more mass/speed there were higher values in the table.

This means that my hypothesis at the beginning was correct, more of m/s means KE will increase proportionally because they are all linear. I found it surprising when the bean bag height went over the water bottle drop mark.

4. To conclude, my hypothesis matched my data. The data values went up when more mass or speed was added. This means if I were in a situation where I needed more kinetic energy for something, I would know to increase mass or the speed of the object giving it energy.

The reason that this hypothesis is correct is when you have more mass, you have more energy. So, when you drop let's say a baseball, it isn’t that heavy so it would only launch the bean bag so far. But a bowling ball is very heavy and has lots of energy when falling because of that, it would make the bean bag go very high.

To make this experiment better, I would use a smoother material for the lever so energy wouldn’t be lost by friction from wood rubbing together. Also, maybe a scanner or video camera to more accurately record how far the bean bag went. All of these would help the lab get more precise results, maybe they could be used in a future lab.

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3 years ago