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Nikolay [14]
3 years ago
6

Which formula can be used to calculate the theoretical yield?

Chemistry
1 answer:
Bas_tet [7]3 years ago
5 0

Answer:

Option C, (Actual yield ÷ percent yield) × 100

Explanation:

Theoretical yield is defined as the total amount of product formed for given reactants in a chemical reaction.  It is an ideal case which assumes no exceptions or wastage.

The mathematical relation between the actual yield, percent yield and theoretical yield is as follows -

P.Y. = \frac{M_{A.Y.}}{M_{T.Y.}} * 100

Where

P.Y. represents the percent yield a

M A.Y. represents the mass obtained from actual yield

M T.Y. represents the mass obtained from theoretical yield

Hence, if we rearrange the formula, we get -

M_{T.Y.} = \frac{M_{A.Y.}}{P.Y.} * 100

Hence, option C is correct

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Write the general electron configuration of a transition element(a) in Period 5;
atroni [7]

Electron configurations of  transition elements in Period 5:

Y     Yttrium  [Kr] 4d1 5s2

Zr Zirconium        [Kr] 4d2 5s2

Nb Niobium  [Kr] 4d4 5s1 (*)

Mo Molybdenum [Kr] 4d5 5s1 (*)

Tc Technetium [Kr] 4d5 5s2

Ru Ruthenium [Kr] 4d7 5s1 (*)

Rh Rhodium  [Kr] 4d8 5s1 (*)

Pd Palladium [Kr] 4d10 (*)

Ag Silver  [Kr] 4d10 5s1 (*)

Cd Cadmium  [Kr] 4d10 5s2

What are transition elements?

  • The transition elements or transition metals occupy the short columns in the center of the periodic table, between Group 2A and Group 3A.  
  • They are sometimes called the d-block elements, since in this region the d-orbitals are being filled in, and are also referred to as B-group elements since in most numbering systems of the columns on the periodic table the numerals of these groups are followed by the letter B.

What are period 5 transition elements?

The period 5 transition metals are yttrium (Y), zirconium (Zr), niobium (Nb), molybdenum (Mo), technetium (Tc), ruthenium (Ru), rhodium (Rh), palladium (Pd), silver (Ag), and cadmium (Cd).

To know more about transition elements, refer:

brainly.com/question/2426896

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7 0
2 years ago
What kind of air pressure would you find in Bridgeport, Connecticut?
klasskru [66]

Answer:

Warm

Explanation: because it would be less hot air population

3 0
4 years ago
The distance between the earth and the sun is 90.0 million miles. Calculate this distance in meters up to three significant digi
sammy [17]
Multiply 90 by 1,609.344: 144,840.96 meters :)
8 0
3 years ago
Read 2 more answers
El superóxido de potasio, KO2, se emplea en máscaras de respiración para generar
Orlov [11]

Answer:

Reactivo límite: Superóxido de potasio.

Moles de oxígeno producidas: n_{O_2}=0.11molO_2

Explanation:

Hola,

En este caso, considerando la reacción química llevada a cabo:

4KO_2(s) + 2H_2O(l) \rightarrow 4KOH(s) + 3O_2(g)

Es posible identificar el reactivo límite calculando las moles de superóxido de potasio que serían consumidas por 0.10 mol de agua por medio de la relación molar 4 a 2 que hay entre ellos:

n_{KO_2}^{consumido\ por\ agua}=0.10molH_2O*\frac{4molKO_2}{2molH_2O} =0.2molKO_2

Así, dado que solo hay 0.15 mol the superóxido de potasio, podemos decir que este es el reactivo límite. Luego, calculamos las moles de oxígeno producidas, considerando la relación molar 4 a 3 que hay entre el superóxido y el oxígeno:

n_{O_2}=0.15molKO_2*\frac{3molO_2}{4molKO_2} \\\\n_{O_2}=0.11molO_2

Best regards.

4 0
3 years ago
Rank the following gases in order of decreasing rate of effusion. Rank from the highest to lowest effusion rate. To rank items a
Alinara [238K]

Answer:

H2, He, PH3, F2, C5H10

Explanation:

From Graham's law, we understood that lighter gas will diffuse faster than heavier gas under same condition. Graham's law of diffusion states as follow:

The rate (R) of diffusion of a gas is inversely proportional to the square root of its density (D) provided temperature and pressure remains constant. Mathematically, it is represented as:

R & 1/√D

Recall:

Molar Mass (M) = 2 x vapour density (D)

M = 2D

R & 1/√M

From the above, we can see that the rate is inversely proportional to the square root of the molar mass of substance. This implies that lighter gas will diffuse faster.

Now, to rank the above from the highest to rate rate of effusion, let us determine the molecular weight of each substance. This is illustrated below:

Molar Mass of F2 = 19 x 2 = 38g/mol

Molar Mass of He = 4g/mol

Molar Mass of C5H10 = (12X5) + (10X1) = 70g/mol

Molar Mass of H2 = 2x1 = 2g/mol

Molar Mass of PH3 = 31 + (3x1) = 34g/mol

Now, we can rank the substance beginning from the highest to the lowest rate of effusion as follow:

Substance >> Molar Mass >> Rank

H2 >>>>>>>>> 2g/mol >>>>>> 1

He >>>>>>>>> 4g/mol >>>>>> 2

PH3 >>>>>>>> 34g/mol >>>>> 3

F2 >>>>>>>>>> 38g/mol >>>> 4

C5H10 >>>>>> 70g/mol >>>> 5

3 0
3 years ago
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