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3 years ago

If an atom has an electron configuration of 2.8.2, what will it be most likely to bond with?

Chemistry

1 answer:

Kipish [7]3 years ago

3 0

Answer:

B. Two chlorine atoms

Explanation:

This electronic configuration shows that the given atom is magnesium.

Electronic configuration of magnesium:

Mg¹² = 1s² 2s² 2p⁶ 3s²

There are two valance electrons of magnesium that's why it would react with two atoms of chlorine. Chlorine is present in seventeen group. It has seven valance electrons. It required just one electron to get complete octet. While magnesium needed to lose two electrons to get complete octet. That's why two chlorine atoms bonded with one magnesium atom. Thus both would get complete octet.

Mg + Cl₂ → MgCl₂

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"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"

Salsk061 [2.6K]

Answer:- The Ka for the acid is $1.53*10^-^5$ .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

$Ka = [H^+][A^-]\frac{1}{HA}$

Where, Ka is the acid ionization constant. Let's plug in the values.

$Ka = \frac{X^2}{0.25-X}$

Let's calculate the value of X first using the equation:

$pH = -log[H^+]$ [/tex]

on taking antilog ob above equation we get:

$[H^+]=10^-^p^H$

$[H^+]=10^-^2^.^7^1$

$[H^+]$ = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

$Ka=\frac{(0.00195)^2}{0.25-0.00195}$

$Ka=1.53*10^-^5$

So, the value of Ka for butyric acid is $1.53*10^-^5$ .

8 0

3 years ago

Read 2 more answers

Using the following equation: 3H2SO4+ 2Fe -> Fe2(SO4)3+ 3H2

Mariana [72]

Answer:

H₂SO₄

Explanation:

Given data:

Number of moles of H₂SO₄ = 15 mol

Number of moles of Fe = 13 mol

Which reactant is limiting reactant = ?

Solution:

Chemical equation:

3H₂SO₄ + 2Fe → Fe₂(SO₄)₃ + 3H₂

now we will compare the moles reactant with product.

H₂SO₄ : Fe₂(SO₄)₃

3 : 1

15 : 1/3×15 = 5

H₂SO₄ : H₂

3 : 3

15 : 15

Fe : Fe₂(SO₄)₃

2 : 1

13 : 1/2×13 = 6.5

Fe : H₂

2 : 3

13 : 3/2×13 = 19.5

Number of moles of product formed by H₂SO₄ are less thus it will act as limiting reactant.

8 0

3 years ago

PLEASE ANSWER BOTH QUESTIONS I WILL MARK YOU BRAINLIEST IF IT IS CORRECT!!

seropon [69]

Answer:

1st Question: A

2nd Question: B

Explanation:

The 1st answer would be A because if a sample is at absolute zero then the sample is at its lowest temperature none of the molecules would be able to move, this is because lower temperature= lower kinetic energy.

The 2nd answer would be B because if a sample has more temperature it speeds up it has more temperature and more kinetic energy, meaning it would move faster because there is more temperature.

7 0

3 years ago

The concentration of carbon monoxide (CO), a common air pollutant, is found in a room to be 5.7 x 10^-3 mg/cm^3. How many grams

damaskus [11]

The amount, in mg, of CO present in the room will be 191,520 mg.

<h3>Stoichiometric problem</h3>

The concentration of the gas in the room is 5.7 x $10^{-3}$ mg/cm3.

The dimension of the room is 3.5 m x 3.0 m x 3.2 m. This is equivalent to 350 cm x 300 cm x 320 cm.

We can obtain the volume of the room as:

350 x 300 x 320 = 33,600,000 cm3

The concentration is in mg/cm3, meaning that it is mass/volume.

Thus:

mass = concentration x volume = 5.7 x $10^{-3}$ mg/cm3 x 33,600,000 cm3

= 191,520 mg

The mass of CO in the room is 191,520 mg

More on stoichiometric problems can be found here: brainly.com/question/14465605

#SPJ1

5 0

1 year ago

A student makes a mistake while preparing a vitamin C sample for titration and adds the potassium iodide solution twice. How wil

LiRa [457]

The added KI does not have any impact

The reaction invovles Titration of vitaminc ( Ascorbic acid)

ascorbic acid + I₂ → 2 I⁻ + dehydroascorbic acid

the excess iodine is free reacts with the starch indicator, forming the blue-black starch-iodine complex.

This is the endpoint of the titration. since alreay excess KI is added ( the source of Iodine), it does not have an influence.

Answer B

Hope this helps!

5 0

4 years ago