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sveticcg [70]
3 years ago
7

Identify the state(s) of matter that each property describes.

Chemistry
1 answer:
disa [49]3 years ago
8 0

Answer:

See below!

Explanation:

Gases and liquids take the shape of the container.

A gas fills all available space.

A solid maintains its shape.

A liquid can be poured.

A gas is compressible.

A liquid and solid have fixed volumes.

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Write the possible sets of quantum numbers of l=3
kykrilka [37]

Answer:

Explanation:

There are four quantum numbers:

Principal quantum number (n)

Azimuthal quantum number (l)

Magnetic quantum number (ml)

Spin quantum number (ms)

All these four quantum numbers gives complete information about an electron like its spin, shells, subshells and orbitals.

For example:

If l=3 than possible sets of quantum numbers are:

Azimuthal quantum number (l)

The azimuthal quantum number describe the shape of orbitals. Its value for s, p, d, f... are 0, 1, 2, 3. For l=3

(n-1)

4-1 = 3

it means principle quantum is 4 and electron is present in f subshell.

Magnetic quantum number (ml)

It describe the orientation of orbitals. Its values are -l to +l. For l=3  the ml will be -3 -2 -1 0 +1 +2 +3.

Spin quantum number (ms)

The spin quantum  number tells the spin of electron either its clock wise (+1/2) or anti clock wise (-1/2).

If the electron is added in full empty orbital its spin will be +1/2 because it occupy full empty. If electron is already present and another electron is added then its spin will be -1/2.

4 0
3 years ago
Structure of tetrazine
Vika [28.1K]

Answer:

Tetrazine is a compound that consists of a six-membered aromatic ring containing four nitrogen atoms with the molecular formula C2H2N4.

(See the image)

Hope it helps!

3 0
3 years ago
A gas system contains 2.00 moles of O2 and CO2 gas, has an initial temperature of 25.0 oC and is under 1.00 atm of pressure. If
Ainat [17]

Answer:

V_2=52.2L

Explanation:

Hello there!

In this case, by bearing to to mind the given conditions, it is firstly possible to determine the initial volume of the closed system via the ideal gas equation:

PV=nRT\\\\V=\frac{2.00mol*0.08206\frac{atm*L}{mol*K}*298.15K}{1.00 atm} \\\\V=48.9L

Which is V1 in the Charles' law:

\frac{T_2}{V_2} =\frac{T_1}{V_1}

And of course, T1 is 298.15 (25+273.15). Therefore, by solving for V2 as the final volume, we obtain:

V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{48.9L*(45+273.15)K}{(25+273.15)K}  \\\\V_2=52.2L

Best regards!

8 0
3 years ago
Picture of gas laws question below:
Vera_Pavlovna [14]

Volume of the tank is 5.5 litres.

Explanation:

mass of the CO2 is given 8.6 grams

Pressure of the gas is 89 Kilopascal which is 0.8762 atm

Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)

R = gas constant 0.0821 liter atmosphere per kelvin)

FROM THE IDEAL GAS LAW

PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)

no of moles = mass/atomic mass

                    =  8.6/44

                    = 0.195 moles

now putting the values in equation

V=nRT/P

  = 0.195*0.0821*302/ 0.8762

  = 5.5 litres.

As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.

4 0
3 years ago
Which of the following acids should be used to prepare a buffer with a pH of 4.5?A. HOC6H4OCOOH, Ka = 1.0 x 10^-3B. C6H4(COOH)2,
ioda

Answer:

C. CH3COOH, Ka = 1.8 E-5

Explanation:

analyzing the pKa of the given acids:

∴ pKa = - Log Ka

A. pKa = - Log (1.0 E-3 ) = 3

B. pKa = - Log (2.9 E-4) = 3.54

C. pKa = - Log (1.8 E-5) = 4.745

D. pKa = - Log (4.0 E-6) = 5.397

E. pKa = - Log (2.3 E-9) = 8.638

We choose the (C) acid since its pKa close to the expected pH.

⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):

  • pH = pKa + Log ([CH3COO-]/[CH3COOH])

∴ pH = 4.5

∴ pKa = 4.745

⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])

⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])

⇒ 0.5692 = [CH3COO-]/[CH3COOH]

∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]

⇒ 1.8 E-5 = [H3O+](0.5692)

⇒ [H3O+] = 3.1623 E-5 M

⇒ pH = - Log ( 3.1623 E-5 ) = 4.5

6 0
3 years ago
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