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3 years ago

When am atom loses electron it becomes

1 answer:

4 0

It becomes an ion.

Orginally, the number of electrons of an atom should be same as the number of protons, and the atom should be electrically neutral, as each proton brings a positive charge and each electron brings a negative charge. However, when an atom loses or gains electrons, the charge would be positive or negative, depending on what the change is. In this way, the atom is no longer an atom, but an ion.

The number of electrons gained or lose by an atom depends on their group number in the periodic table. They'll gain or loose the least electrons in order to have an octet of electrons (or duplet). Which means, their outermost shell electrons would be 8, or 2 for the first shell.

One exam is fluorine. It is in the group VII, it has an outer most shell electron of 7. In order to make fluoride ion, one electron is gained, so that the outer most shell will have 8 electrons. Therefore, fluoride ion has a negative charge, as one electron brings a negative charge.

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Which of the following is NOT a reason for the experimental volume of the flask to be incorrect?

dsp73

Answer:

Explanation:

i dont andesdant

4 0

3 years ago

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

3 0

3 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

6 0

3 years ago

A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o

HACTEHA [7]

Answer:

1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Explanation:

From the question given above, the following were obtained:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Volume of solution = 20.4 mL

Concentration of solution =?

Next, we shall determine the mass of sodium hydroxide, NaOH. This can be obtained as as illustrated below:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Mass of NaOH =?

Mass of NaOH = (Mass Cylinder + NaOH) – (Mass of empty cylinder)

Mass of NaOH = 31.92 – 9.5

Mass of NaOH = 22.42 g

Finally, we shall determine concentration of the solution as follow:

Mass of NaOH = 22.42 g

Volume of solution = 20.4 mL

Concentration of solution =?

Concentration = mass /volume

Concentration of solution = 22.42 / 20.4

Concentration of solution = 1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Therefore, the concentration of the solution is 1.1 gmL¯¹

3 0

3 years ago

Calculate the mass of .0605 moles of crystalline sucrose. Sucrose is C12 H22 O11 . The molar mass is __________ g/mole. Show you

KengaRu [80]

Answer:

20.7g/mol

Explanation:

7 0

3 years ago