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Lynna [10]
2 years ago
14

A laboratory analysis of an unknown sample yields 74.0% carbon, 7.4% hydrogen, 8.6% nitrogen, and 10.0% oxygen. What is the empi

rical formula of the sample? Give your answer in the form C#H#N#O# where the number following the element’s symbol corresponds to the subscript in the formula. (Don’t include a 1 subscript explicitly). For example, the formula CH 22 O would be entered as CH2O.
Chemistry
1 answer:
kirza4 [7]2 years ago
3 0
Let us assume that there is a 100g sample present. The respective mass of each element will then be:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Now, we divide each constituent's mass by its Mr to obtain the moles of each
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Dividing by the smallest number:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula is
C10H12NO
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PLS HELP ASAP, WILL GIVE 100 POINTS.
Ivan

Answer: values of the quantum numbers: -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6

location of the electron: In the 7th energy level away from the nucleus.

Explanation:

From the description of the problem, the magnetic number is given is as -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6 and the electron is located in the 7th energy level away from the nucleus. Basically, the problem is testing for the understanding of the principal quantum numbers which gives the location of electrons and the magnetic quantum number that shows the spatial orientation of the orbitals.

           The orbital designation of the describe electron is 7d

Magnetic quantum number is limited by the azimuthal quantum number which is the quantum number describing the possible shapes. The azimuthal is given as L= n-1. "n" is the principal quantum number which is 7. Therefore L is 6 and the magnetic quantum numbers are -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6

The position of the electron is given by the principal quantum number which represents the main energy level in which the orbital is located or the average distance from the nucleus.

Got it from quasarJose

Hope it helps

5 0
2 years ago
What are the 3 critical components of an electromagnet and what purpose do they each serve?
WARRIOR [948]

Answer:

A source of electricity, a wire coil, and an iron core  

Explanation:

An electromagnet has three critical components:

1. A source of electricity

This is often a battery.

It generates the electric current that produces the magnetic field.  

2. A wire coil

The wire carries the electric current.

Stacking the wire into loops makes a stronger magnetic field.

The more loops in the coil, the stronger the field.

3. An iron core

An iron core greatly increases the strength of the magnetic field within it and at its ends.

4 0
2 years ago
Item 5 A solution of methanol, CH3OH, in water is prepared by mixing together 128 g of methanol and 108 g of water. The mole fra
Basile [38]

Answer:

Mole fraction of methanol will be closest to 4.

Explanation:

Given, Mass of methanol = 128 g

Molar mass of methanol = 32.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{128\ g}{32.04\ g/mol}

Moles\ of\ methanol = 3.995\ mol

Given, Mass of water = 108 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{108\ g}{18.0153\ g/mol}

Moles\ of\ water= 5.995\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ methanol=\frac {n_{methanol}}{n_{methanol}+n_{water}}

Mole\ fraction\ of\ methanol=\frac{3.995}{3.995+5.995}=0.39989

<u>Mole fraction of methanol will be closest to 4.</u>

5 0
2 years ago
Octane is a liquid component of gasoline. Given the following vapor pressures of octane at various temperatures, estimate the bo
Hitman42 [59]

Answer:

110.8 ºC

Explanation:

To solve this problem we will make use of the Clausius-Clayperon equation:

lnP = - ΔHºvap/RT + C

where P is the pressure, ΔHºvap is the enthalpy of vaporization, R is the gas constant, T is the temperature, and C is a constant of integration.

Now this equation has a form y = mx + b where

y = lnP

x = 1/T

m = -ΔHºvap/R

Now we have to assume that ΔHºvap remains constant which is a good asumption given the narrow range of temperatures in the data ( 104-125) ºC

Thus what we have to do is find the equation of the best fit for this data using a  software as excel or your calculator.

T ( K)               1/T                  ln P

377               0.002653       5.9915

384              0.002604       6.2115

390              0.002564       6.3969

395              0.002532       6.5511

398              0.002513        6.6333

The best line has a fit:

y = -4609.5 x  + 18.218

with R² = 0.9998

Now that we have the equation of the line, we simply will substitute for a pressure of 496 mm in Leadville.

ln(496) = -4609.5(1/Tb) + 18.218

6.2066 = -4609.5(1/Tb) +18.218

⇒ 1/Tb = (18.218 - 6.2066)/4609.5 = 0.00261

Tb = 383.76 K  = (383.76 -273)K = 110.8 ºC

Notice we have touse up to 4 decimal places since rounding could lead to an erroneous answer ( i.e boiling temperature greater than 111, an impossibility given the data in the question). This is as a result of the value 496 mmHg so close to 500 mm Hg.

Perhaps that is the reason the question was flagged.

7 0
2 years ago
When 28.0 g of acetylene reacts with hydrogen, 24.5 g of ethane is produced. What is the percent yield of C2H6 for the reaction?
Afina-wow [57]

Answer:

Y=75.6\%

Explanation:

Hello.

In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:

m_{C_2H_6}^{theoretical}=28.0gC_2H_2*\frac{1molC_2H_2}{26gC_2H_2}*\frac{1molC_2H_6}{1molC_2H_2}  *\frac{30gC_2H_6}{1molC_2H_6}\\ \\m_{C_2H_6}^{theoretical}=32.3gC_2H_6

Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:

Y=\frac{24.5g}{32.3g}*100\%\\ \\Y=75.6\%

Best regards.

3 0
2 years ago
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