Select the correct answer.

How does a negative ion differ from a positive?

Ilya [14]

A negative ion or anion is an electrically charged atom or group of atoms that are formed by gaining one or more electrons. Whereas a positive ion or cation is an electrically charged atom or group of atoms that are formed by losing one or more electrons.
The losing and gaining all occurs on the outermost shell.

7 0

3 years ago

What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?

Ilia_Sergeevich [38]

Answer: The volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of $O_{2}$ gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of $O_{2}$ (molar mass = 32.0 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol$

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L$

Thus, we can conclude that the volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

6 0

3 years ago

These models show the electron structures of two different nonmetal elements. Element 1 at left has a purple circle at center wi

Daniel [21]

Answer:

Element 2

Explanation:

If we look at the model stated for element 1, it is clear that element 1 must be a noble gas. It has eight electrons in its outermost shell this implies that it has already attained a complete octet of electrons and is reluctant towards chemical reaction.

The second element belongs to group 16 since it has six electrons on its outermost shell. It is certainly more reactive than element 1 which is a noble gas.

7 0

3 years ago

Read 2 more answers

How could measuring the melting point of a solid help decide whether it was a mixture or a compound?

mars1129 [50]

Answer: It can't.

Explanation:

In most cases, the melting point alone will not enable you to identify a compound. Millions of solid organic compounds, and their melting points, are known. Perhaps 10,000 of these will have the same melting point as your unknown compound.

Hope this helps!

8 0

3 years ago

An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To

antoniya [11.8K]

Answer: The new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

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

The expression of $K_p$ for above reaction follows:

$K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}$ ........(1)

We are given:

$P_{PCl_5}=217.0torr$

$P_{PCl_3}=13.2torr$

$P_{Cl_2}=13.2torr$

Putting values in above equation, we get:

$K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24$

Now we have to calculate the new partial pressure of $Cl_2$ .

$P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}$

$217.0torr+13.2torr+P_{Cl_2}=263.0torr$

$P_{Cl_2}=32.8torr$

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of $Cl_2$ .

Now, the equilibrium is shifting to the reactant side. The equation follows:

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

Initial: 13.2 32.8 217.0

At eqm: 13.2-x 32.8-x 217.0+x

Putting values in expression 1, we get:

$1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38$

Neglecting the 40.4 value of 'x' because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of $PCl_5$ = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of $PCl_3$ = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of $Cl_2$ = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0

3 years ago