All categories

3 years ago

Which represents the ionization of a strong eletrolyte

1 answer:

3 0

Answer: The answer is K3PO4(s) → 3K+(aq) + PO43–(aq) since water-soluble ionic tripotassium phosphate dissociates completely into K+ and PO43– ions when dissolved, that is, no K3PO4 remains in the solution. Carbonic acid H2CO3 and acetic acid CH3COOH are weak electrolytes since they are weak acids that do not completely ionize, while nonelectrolyte CH3OH do not dissociate into ions.

You might be interested in

(this is science not chemistry)

erma4kov [3.2K]

I'm pretty sure ita A. acceleration

4 0

3 years ago

Read 2 more answers

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or

Ahat [919]

Answer :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

Rules for Oxidation Numbers are :

The oxidation number of a free element is always zero.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Now we have to determine the oxidation state of the elements in the compound.

(a) $H_2SO_4$

Let the oxidation state of 'S' be, 'x'

$2(+1)+x+4(-2)=0\\\\x=+6$

Hence, the oxidation state of 'S' is, (+6)

(b) $Ca(OH)_2$

Let the oxidation state of 'Ca' be, 'x'

$x+2(-2+1)=0\\\\x=+2$

Hence, the oxidation state of 'Ca' is, (+2)

(c) $BrOH$

Let the oxidation state of 'Br' be, 'x'

$x+(-2)+1=0\\\\x=+1$

Hence, the oxidation state of 'Br' is, (+1)

(d) $ClNO_2$

Let the oxidation state of 'N' be, 'x'

$-1+x+2(-2)=0\\\\x=+5$

Hence, the oxidation state of 'N' is, (+5)

(e) $TiCl_4$

Let the oxidation state of 'Ti' be, 'x'

$x+4(-1)=0\\\\x=+4$

Hence, the oxidation state of 'Ti' is, (+4)

(f) $NaH$

Let the oxidation state of 'Na' be, 'x'

$x+(-1)=0\\\\x=+1$

Hence, the oxidation state of 'Na' is, (+1)

4 0

3 years ago

For the reaction;

ankoles [38]

Answer:

At equilibrium:

[H2] = 0.005 M

[Br2] = 0.105 M

[HBr] = 0.189 M

Explanation:

H2(g) + Br2(g) ⇄ 2HBr

an "x" value will be used from reactant to produced "2x"

so at equilibrium:

[H2] = 0.1 - x

[Br2] = 0.2 - x

[HBr] = 2x

we know that Kc=[HBr]²/[H2][Br2]

Thus 62.5 = (2x)²/(0.1-x)(0.2-x)

this generate a quadratic equation: 58.5x² - 18.75x + 1.25 = 0

the x₁ = 0.23 x₂ = 0.09457

we pick 0.09457 because the two reactants can not make more than what they have. x₁ is higher than both initial reactant concentration

Then we substitute the "x₂" value at equilibrium:

[H2] = 0.1-0.09457 = 0.005 M

[Br2] = 0.2-0.09457 = 0.105 M

[HBr] = 2*0.09457 = 0.189 M

7 0

4 years ago

5. The statement that in undisturbed sedimentary rock layers,

Colt1911 [192]

Answer:

The correct answer is - option C. Law of superposition.

Explanation:

According to the law of superposition states that within a sequence of layers of natural sedimentary rock without any disturbance, the oldest layer is lies under the younger layers with ascending order in the sequence.

If a sedimentary rock has 4 layers as a, b, c and d where the younger layer d would be on top and the oldest layer a would be at the bottoms.

Thus, the correct answer is - option C.

7 0

3 years ago

Read 2 more answers

Analysis of a compound indicates that it contians 1.04 g K, 0.70 g Cr, and 0.86 g O. Find its empirical formula.

Annette [7]

Answer:

K2CrO4

Explanation:

To find the empirical formula, we need to divide each element by its atomic mass. The atomic masses of potassium, chromium and oxygen are 39.0983, 51.9961 and 15.999 respectively. We make the divisions as follows:

K = 1.04/39.0983 = 0.027

Cr = 0.70/51.9961 = 0.013

O = 0.86/15.999 = 0.054

We now divide by the smallest which is the number of moles of the Chromium

K = 0.027/0.013 = 2

Cr = 0.013/0.013 = 1

K = 0.054/0.013 = 4 approximately

The empirical formula is thus:

K2CrO4

7 0

3 years ago