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mafiozo [28]
3 years ago
7

How many acetate ions are formed when three formula units of zinc acetate dissociate?

Chemistry
1 answer:
jeyben [28]3 years ago
3 0
A formula unit is the same as the empirical formula of a compound or an ionic molecule. It is the lowest ratio of the atoms in the compound or ion. Zinc acetate ions dissociates into zinc ions and acetate ions. The dissociation reaction is expressed as follows:
Zn(O2CCH3)2 = Zn2+ + 2(O2CCH3)1-

We determine the amount of acetate ions produced as follows:

Moles Zn(O2CCH3)2 = (3 formula units Zn(O2CCH3)2) ( 1 mol / 6.022x10^23 formula units) = 4.98x10^-24 mol Zn(O2CCH3)2
moles (O2CCH3)1- = 4.98x10^-24 mol Zn(O2CCH3)2 ( 2 mol (O2CCH3)1- / 1 mol Zn(O2CCH3)2 ) = 9.96x10^-24 mol (O2CCH3)1-
# of acetate ions = 9.96x10^-24 mol (O2CCH3)1- ( 6.022x10^23 ions / 1 mol (O2CCH3)1-) = 6 acetate ions
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mario62 [17]

Answer:

  • <u><em>a. C₂H₄</em></u>

Explanation:

At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).

Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.

Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).

A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.

<u>a. C₂H₄:</u>

  • C₂H₄ (g) + 3O₂ (g) → 2CO₂(g)  + 2H₂O (g)

Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.

The following analysis just shows that the other options are not right.

<u>b. C₂H₂:</u>

  • 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g)  + 2H₂O (g)

The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.

<u>с. С₃Н₈</u>

  • C₃H₈ (g) + 5O₂ (g) → 3CO₂(g)  + 4H₂O (g)

The mole ratio is 1 mol C₃H₈ : 5 mol O₂

<u>d. C₂H₆</u>

  • 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g)  + 6H₂O (g)

The mole ratio is 2 mol C₂H₆ : 7 mol O₂

7 0
3 years ago
A strong acid is one that:
HACTEHA [7]

Answer:

A strong acid completely ionizes in water.

5 0
3 years ago
Read 2 more answers
An analysis of a sugar compound is found to contain 132 grams of carbon (C), 22 grams of hydrogen (H), and 176 grams of oxygen (
ddd [48]
132 g of C  ,   22 g of H   , 176 g of O

132 + 22 + 176 => 330 g <span>of the substance

</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u        H = 1.0 u       O = 16.0 u

C = 132 / 12.0 => 11 moles

H = 22 / 1.0 => 22 moles

O = 176 / 16.0 => 11 moles

Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1

H = 22 / 11 => 2

O = 11 / 11 => 1

 formula empirically <span>is : CH</span>₂O

hope this helps!


8 0
3 years ago
Calculate the number of moles of a gas that is present in a 7.55 L container at 45°C, if the gas exerts a pressure of 725mm Hg.
vovikov84 [41]

<u>Answer:</u> The number of moles of gas present is 0.276 moles

<u>Explanation:</u>

To calculate the number of moles of gas, we use the equation given by ideal gas:

PV = nRT

where,

P = Pressure of the gas = 725 mm Hg

V = Volume of the gas = 7.55 L

n = number of moles of gas = ?

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of the gas = 45^oC=(45+273)K=318K

Putting values in above equation, we get:

725mmHg\times 7.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 318K\\\\n=0.276mol

Hence, the number of moles of gas present is 0.276 moles

4 0
3 years ago
A certain gas is present in a 10.0 LL cylinder at 4.0 atmatm pressure. If the pressure is increased to 8.0 atmatm the volume of
ICE Princess25 [194]

Answer:

The gas obeys Boyle’s law and the value of k_i\&k_f both are equal to 40.0 atm L.

Explanation:

Initial volume of the gas = V_1=10.0 L

Initial pressure of the gas = P_1=4.0 atm

Final volume of the gas = V_2=5.0 L

Final pressure of the gas = P_2=8.0 atm

This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

PV=k

The equation given by this law is:

P_1V_1=P_2V_2

P_1\propto \frac{1}{V_1}

P_1V_1=k_i

k_i=4.0 atm\times 10.0 L = 40.0 atm L

P_2\propto \frac{1}{V_2}

P_V_2=k_f

k_f=8.0 atm\times 5.0 L = 40.0 atm L

k_i=k_f=40.0 atm L

The gas in the cylinder is obeying Boyle's law.

The gas obeys Boyle’s law and the value of k_i\&k_f both are equal to 40.0 atm L.

6 0
3 years ago
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