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Archy [21]
2 years ago
6

If you produce 14.5 g of O2 in the lab from a reaction that you showed by calculation could produce a maximum of 22.0 g of O2, w

hat is your percent yield?
Chemistry
2 answers:
Komok [63]2 years ago
5 0

Percent yield is simply the ratio of the actual amount produced over the theoretical amount that can be produced. In this case that would be:

percent yield = actual amount / theoretical amount

percent yield = (14.5 g / 22.0 g) * 100%

<span>percent yield = 65.91%</span>

almond37 [142]2 years ago
5 0

Answer: 65.9 %

Explanation:

The formula for calculating percentage yield is:

Percentage yield = \frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Given : Experimental yield = 14.5 g

Theoretical yield = 22.0 g

Putting in the values we get:

Percentage yield = \frac{14.5}{22.0}\times 100=65.9\%

Therefore, the percentage yield is 65.9 %.

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2 years ago
Dry ice is solid carbon dioxide. A 0.050-g sample of dry ice is placed in an evacuated 4.6-L vessel at 30 °C. Calculate the pres
goldenfox [79]

The answer is 6.1*10^-3 atm.

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3 0
2 years ago
0.15 gm of metallic oxide was dissolved in 100 ml of 0.1 N H2SO4 and 25.8 ml of 0.095N NaOH were used to neutralise the remainin
Mila [183]

Answer:

Explanation:

25.8 ml of .095 N NaOH is needed to neutralise the remaining acid

equivalent of NaOH used = 25.8 x .095 / 1000 = .002451 gm equivalent .

acid remaining = .002451 gm equivalent .

acid initially taken = 100 ml of .1 N  /  1000  = . 01 gm equivalent

acid reacted with metal = .01  -.002451 = .007549 gm equivalent

This must have reacted with same gram equivalent of metal oxide

.007549  gm equivalent = .15 gm of metal oxide

1 gm equivalent = 19.87 gm

equivalent weight of metal = 19.87 - equivalent weight of oxygen

= 19.87 - 8 = 11.87 .

1

7 0
2 years ago
(a) What are the possible values of l for n = 4? (Enter your answers as a comma-separated list.)
vodka [1.7K]

Answer:

(a) 0,1,2,3      (b) -3,-2,-1,0,1,2,3              (c) 6               (d) 5

Explanation:

(a) for the principal quantum number 'n', the possible values of I = 0 to n-1. Thus, if the principal quantum number 'n' =4, I = 0,1,2,3.

(b) for a given number of 'I', the possible values of ml = -I to +I. Therefore, if I =3, then ml = -3,-2,-1,0,1,2,3

(c) 'I' which is the orbital angular momentum quantum number usually has values from 0,1,2,⋯,n−1. Therefore, for n greater than or 6, t would be greater than or equal to 5. Thus, the smallest possible value of n for which I can be 6 is 6.

(d) In a 3-dimensional figure,  If the z-component of the orbital angular momentum Lz for which I=5 is measured, The possible outcomes will be:

mħ = -5ħ, -4ħ, -3ħ, -2ħ, -1ħ, 0, 1ħ, 2ħ, 3ħ, 4ħ, 5ħ.

Thus, the smallest possible l that can have a z component of 5ℏ is 5.

4 0
2 years ago
Select the correct set of quantum numbers (n, l, ml, ms) for the first electron removed in the formation of a cation for stronti
matrenka [14]

Answer:

5,0,0,-1/2

Explanation:

The quantum numbers are a way to characterize the electrons, and so, identify the region that it's more probable to find it (orbital). They are:

- Principal quantum number (n): represents the shell or level, and varies from 1 to 7, and are represented by the letter K, L, M, N, O, P, and Q.

- Azimuthal quantum number (l): represents the subshell or sublevel, and is represented by 0,1,2,3.., and for the letters s, p, d, f,...

- Magnetic quantum number (ml): represents the orbital. It varies from -l to +l passing by 0. Each orbital can have 2 electrons.

- Spin quantum number (ms): represents the spin of the electron. It can be +1/2 or -1/2.

The strontium has an atomic number equal to 38, by the Linus Pauling's diagram, the electronic distribution is:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²

The valence electron is at the subshell 5s, which has only one magnetic quantum number: 0. Because it has 2 electrons, the first one has spin =1/2, and the other -1/2. So the first electron of the formation of cation has quantum numbers:

n = 5; l = 0; ml = 0; ms = -1/2

7 0
1 year ago
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