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3 years ago

If the pressure, volume, and temperature of a gas are known, which can most lik

Chemistry

1 answer:

morpeh [17]3 years ago

5 0

Answer:

The molar amount of the gas.

Explanation:

If the pressure, volume, and temperature of a gas are known than molar amount can be determine.

Formula:

PV = nRT

P = pressure

V = volume

R = general gas constant

T = temperature

n = number of moles

Example:

V = 2 L, P = 1 atm , T = 273 K , R = 0.0821 atm. L/mol.K

n = PV/RT

n = 1 atm. 2 L / 0.0821 atm. L/mol.K . 273 K

n = 2 / 22.413 /mol

n = 0.09 mol

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Given that E°red = -0.40 V for Cd2+/Cd at 25°C, find E° and E for the concentration cell expressed using shorthand notation belo

tensa zangetsu [6.8K]

From the information in the question, the E° and E for the cell is 0.00 V and 0.12 V.

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

We know that E°cell = 0.00 V since the anode and cathode are both made up of cadmium.

Substituting the given values into the Nernst equation;

Ecell = 0.00 V - 0.0592/2 log (1.0 × 10-5 M/0.100 M)

Ecell = 0.00 V - 0.0296 log(1 × 10^-4)

Ecell = 0.12 V

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3 years ago

Complete the statements by writing the number

dezoksy [38]

Answer :

The substance is in the gas phase only in region → 5

The substance is in both the liquid and the solid phase in region → 2

The substance is in only the liquid phase in region → 3

The melting point is the temperature at region → 2

The boiling point is the temperature at region → 4

Explanation :

Six phases of substance:

Melting or fusion : In this process the phase changes from solid state to liquid state at constant temperature.

Freezing : In this process the phase changes from liquid state to solid state at constant temperature.

Evaporation : In this process the phase changes from liquid state to gaseous state at constant temperature.

Condensation : In this process the phase changes from gaseous state to liquid state at constant temperature.

Sublimation : In this process the phase changes from solid state to gaseous state without passing through the liquid state at constant temperature.

Deposition : In this process the phase changes from gaseous state to solid state without passing through the liquid state at constant temperature.

3 0

3 years ago

The group of atoms that is responsible for the characteristic properties of a family of organic compounds is called a/an _______

Sindrei [870]

The group of atoms that is responsible for the characteristic properties of a family of organic compounds is called a <u>functional</u> group.

Functional groups are groups of one or extra atoms with exclusive chemical homes regardless of what's connected to them. The atoms of practical agencies are certain via covalent bonds with one another and with the relaxation of the molecule.

Functional groups include:

hydroxyl,
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carbonyl,
carboxyl,
amino, phosphate,
sulfhydryl.

An atom or institution of atoms that replaces hydrogen in a natural compound that defines the shape of a family of compounds and determines the properties of the family.

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6 0

2 years ago

A solution of Na2CO3 is added dropwise to a solution that contains 1.15×10−2 M Fe2+ and 0.58×10−2 M Cd2+. What concentration of

castortr0y [4]

The question is incomplete, complete question is;

A solution of $Na_2CO_3$ is added dropwise to a solution that contains $1.15\times 10^{-2} M$ of $Fe^{2+}$ and $0.58\times 10^{-2} M$ and $Cd^{2+}$ .

What concentration of $CO_3^{2-}$ is need to initiate precipitation? Neglect any volume changes during the addition.

$K_{sp}$ value $FeCO_3: 2.10\times 10^{-11}$

$K_{sp}$ value $CdCO_3: 1.80\times 10^{-14}$

What concentration of $CO_3^{2-}$ is need to initiate precipitation of the first ion.

Answer:

Cadmium carbonate will precipitate out first.

Concentration of $CO_3^{2-}$ is need to initiate precipitation of the cadmium (II) ion is $3.103\times 10^{-12} M$ .

Explanation:

1) $FeCO_3\rightleftharpoons Fe^{2+}+CO_3^{2-}$

The expression of an solubility product of iron(II) carbonate :

$K_{sp}=[Fe^{2+}][CO_3^{2-}]$

$2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}$

$[CO_3^{2-}]=1.826\times 10^{-9}M$

2) $CdCO_3\rightleftharpoons Cd^{2+}+CO_3^{2-}$

The expression of an solubility product of cadmium(II) carbonate :

$K_{sp}=[Cd^{2+}][CO_3^{2-}]$

$1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}$

$[CO_3^{2-}]=3.103\times 10^{-12} M$

On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.

So, cadmium carbonate will precipitate out first.

And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the $3.103\times 10^{-12} M$ concentration.

6 0

3 years ago

Naturally occurring copper exists in two isotopic forms: 63Cu and 65Cu. The atomic mass of copper is 63.55 amu. What is the appr

Mariulka [41]

To calculate the average atomic weight, each exact atomic weight is multiplied by its percent abundance, then, add the results together. If the natural abundance of 63Cu is assigned x, the natural abundance of 65Cu is 1-x (the two abundance always add up to 1). So the solution is: (63)(x)+(65)(1-x) = 63.55, 63x+65-65x=63.55, x=0.725=72.5%. The natural abundance of 63Cu is 72.5%, that of 65Cu is 1-72.5%=27.5%.

3 0

3 years ago