Which factor is least likely to cause a memory problem

Find the percent ionization of a 0.337 m hf solution. the ka for hf is 3.5 x 10-4. 1.1 % 1.2 x 10-2 % 3.2 % 3.5 x 10-2 % 4.7 %

Ugo [173]

To determine the percent ionization of the acid given, we make use of the acid equilibrium constant (Ka) given. It is the ration of the equilibrium concentrations of the dissociated ions and the acid. The dissociation reaction of the HF acid would be as follows:<span>

HF = H+ + F-

The acid equilibrum constant would be expressed as follows:

Ka = [H+][F-] / [HF] = 3.5 x 10-4

To determine the equilibrium concentrations we use the ICE table,
HF H+ F-
I 0.337 0 0
C -x +x +x
---------------------------------------------
E 0.337-x x x

3.5 x 10-4 = [H+][F-] / [HF]
3.5 x 10-4 = [x][x] / [0.337-x] </span>

Solving for x,

x = 0.01069 = [H+] = [F-]

percent ionization = 0.01069 / 0.337 x 100 = 3.17%

8 0

3 years ago

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Metals (M) in group 2 of the periodic table form ionic bonds

egoroff_w [7]

Answer:

16

Explanation:

Group two elements are alkaline earth metal.

All these have two valance electrons. In order to achieve noble gas configuration it loses its two valance and get complete octet.

Reaction with group 16.

Reaction with oxygen,

They react with oxygen and form oxide.

2Ba + O₂ → 2BaO

2Mg + O₂ → 2MgO

2Ca + O₂ → 2CaO

this oxide form hydroxide when react with water,

BaO + H₂O → Ba(OH)₂

MgO + H₂O → Mg(OH)₂

CaO + H₂O → Ca(OH)₂

With sulfur,

Mg + S → MgS

Ca + S → CaS

Ba + S → BaS

5 0

3 years ago

For each of the following reactions, identify the missing reactant(s) or products(s) and then balance the resulting equation. No

Pachacha [2.7K]

Answer:

A. The reactants are= Li and O2

The balanced equation is:

4Li + O2 —> 2Li2O

B. The products are: MgCl2 and O2

The balanced equation is:

Mg(ClO3)2 —> MgCl2 + 3O2

C. The products are: Ca(NO3)2 and H2O

The balanced equation is:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 +

2H2O

D. The products are: CO2 and H2O

The balanced equation is:

C5H12 + 8O2 —> 5CO2 + 6H2O

Explanation:

A. ____ —> Li2O

The reactants are Li and O2. Thus the equation is given below:

Li + O2 —> Li2O

Thus the equation is balanced as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Li2O as shown below:

Li + O2 —> 2Li2O

Now, we have 4 atoms of Li on the right side and 1 atom on the left. It can be balance by putting 4 in front of Li as shown below:

4Li + O2 —> 2Li2O

Now the equation is balanced

B. Mg(ClO3)2 —> __

The products are MgCl2 and O2

The equation is given below:

Mg(ClO3)2 —> MgCl2 + O2

The equation can be balance as follow:

There are 6 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

Mg(ClO3)2 —> MgCl2 + 3O2

Now the equation is balanced

C. HNO3 + Ca(OH)2 —>___

The products are: Ca(NO3)2 and H2O

The equation is given below:

HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

The equation is balanced as follow:

There are 2 atoms of NO3 on the right side and 1 atom on the left. It can be balance by putting 2 in front of HNO3 as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

There are a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + 2H2O

Now the equation is balanced

D. C5H12 + O2 —>__

The products are: CO2 and H2O

The equation is given below:

C5H12 + O2 —> CO2 + H2O

The equation can be balance as follow:

There are 5 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 5 in front of CO2 as shown below:

C5H12 + O2 —> 5CO2 + H2O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

C5H12 + O2 —> 5CO2 + 6H2O

Now, there are a total of 16 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 8 in front of O2 as shown below:

C5H12 + 8O2 —> 5CO2 + 6H2O

Now we can see that the equation is balanced.

4 0

3 years ago

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What happens to valence electrons when a covalent bond forms?

abruzzese [7]

Explanation:

Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability

5 0

3 years ago

Indicate if the following statements are TRUE (T) or FALSE (F):

Aleks [24]

Answer:

a) T

b) T

c) F

d) F

e) T

f) T

g) T

h) F

I) F

j) F

k) F

l) F

Explanation:

The w/v concentration is obtained from, mass/volume. Hence;

%w/v= 50/1000= 5%

In the %w/w we have;

25g/100 g = 25% w/w

In combustion reaction, energy is given out hence it is exothermic.

Neutralization reaction yields a salt and water

% by mass of carbon is obtained from;

8× 12/114 × 100 = 84.1%

All the ionic substances mentioned have very low solubility in water.

One mole of a substance contains the Avogadro's number of each atom in the compound.

There are two iron atoms so one mole contains 2× 55.85 g of iron.

Some sulphates such as BaSO4 are insoluble in water.

Halides are soluble in water hence NaI is soluble in water.

The equation does not balance with the given coefficients because the number of atoms of each element on both sides differ.

The equation represents a decomposition of calcium carbonate as written.

3 0

3 years ago