1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ghella [55]
2 years ago
13

Select the correct answer.

Chemistry
1 answer:
pochemuha2 years ago
8 0

Answer:

В.

The volume will decrease.

Explanation:

You might be interested in
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
Other questions:
  • Charcoal (burned wood) that was used to make prehistoric drawings on cave walls in france was scraped off and analyzed. the resu
    12·1 answer
  • 1.72 mol of LiCl in 37.5 L of solution
    9·1 answer
  • Earth has one moon and mars has two moons. how many more moons are found around the outer planets than the inner planets?
    12·2 answers
  • he solubility of oxygen gas in water at 25 °C and 1.0 atm pressure of oxygen is 0.041 g/L. The solubility of oxygen in water at
    14·1 answer
  • 50 points! The isotope shown has a mass of 14.003241 amu. Calculate how much energy is released from the binding of 2.530x10 mol
    12·2 answers
  • Will mark brainliest if solved correctly.
    11·2 answers
  • radiowaves in the AM region have frequencies in the range 550 to 1600 kilocycles per second (550 to 1600 kh) calculate the wavel
    11·1 answer
  • 1.
    10·2 answers
  • Thallium-201 has a half-life 73 hours. If 4.0 mg of thallium-201 disintegrates over a period of 5.0 °, how many milligrams of th
    15·1 answer
  • 1. Why do you think pepper behaved the way it did in this experiment? What do you think that sugar and salt would do?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!