CHEMISTRY HIGH SCHOOL

A 1.149 g sample contains only vitamin C (C6H8O6) and sucralose (C12H19Cl3O8). When the sample is dissolved in water to a total volume of 41.1 mL, the osmotic pressure of the solution is 3.18 atm at 287 K. What is the mass percent of vitamin C and sucralose in the sample? a) mass percent vitamin C:%?
b) mass percent sucralose:%?

Answers

Answer 1
Answer:

The mass percentage of vitamin C is 74.5% and the mass percentage of sucralose is 25.5%.

Osmotic pressure (π) = iCRT

Where;

π = Osmotic pressure = 3.18 atm

i = Van't Hoff factor = 1 (molecular solution)

C = concentration (in mol/L)

R = gas constant = 0.082 LatmK-1mol-1

T = temperature =  287 K

But C = number of moles(n) /Volume (V)

volume = 41.1 mL or 0.0411 L

Substituting into the given equation;

π = n/V × RT

Make n the subject of the formula;

n = πV/RT

n = 3.18 atm ×  0.0411 L/0.082 LatmK-1mol-1 × 287 K

n = 0.0056 moles

Mass of vitamin C = a

Mass of sucralose = b

Mass of sample = 1.149 g

Hence;

a + b = 1.149

b = 1.149 - a

Molar mass of vitamin C = 176 g/mol

Molar mass of sucralose = 398 g/mol

Total number of moles = number of moles of vitamin C + number of moles of sucralose

 0.0056  moles = a/176  + (1.149 - a)/398

70048 ( 0.0056 ) = 398a + 176 (1.149 - a)

392.269 = 398a + 202.224 - 176a

392.269 - 202.224 = 398a  - 176a

190.045 = 222a

a = 190.045/222

a = 0.856 g

But;

b = 1.149 - a

Where a =  0.856 g

b = 1.149 - 0.856

b = 0.293 g

Mass percent vitamin C =  0.856 g/1.149 g × 100/1

= 74.5%

Mass percent sucralose = 0.293/1.149 g × 100/1

= 25.5%

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Consider the lewis structures for h2cch2, h2ccch2, and h2cccch2. what hybridization should be expected for the ch2 carbon and the middle ccc carbon, respectively?

Answers

Answer:

CH2 --> sp2 hybridization.

C-C-C --> sp hybridization.

Explanation:

The CH2 has two simple bonds to the H and one double bond to another C. This means that in the hybridization of this C are involved 3 orbitals: 1 s and 2 p, resulting in a sp2 hybridization.

The C-C-C central carbon has two double bonds, one to each C. This means that in the hybridization of this C are involved 2 orbitals: 1 s and 1 p, resulting in a sp hybridization.

Final answer:

The hybridization of the CH2 carbon is 'sp2' and the middle CCC carbon's hybridization in all mentioned molecules is 'sp'.

Explanation:

The carbon in CH2 in all the three molecules, H2CCH2, H2CCCH2, and H2CCCCH2 is always connected to two other atoms: a carbon and a hydrogen. Therefore, its hybridization is

sp2

. The middle carbon in ccc sequence is also linked to two other carbons only. Its hybridization is also

sp

. To summarize, the CH2 carbon's hybridization in all the molecules is

sp2

and the middle CCC carbon's hybridization in all the molecules is

sp

.

Learn more about Hybridization here:

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QUICK QUESTION: On the Bohr model, how come potassium has 19 electrons in its valence shell if potassium has a K+? Isn’t it suppose to have 18 electrons? I thought that if an ion has a positive charge, the atom has lost electrons. Pls help me I WILL GIVE BRAINLIEST TO THE BEST ANSWER

Answers

Answer:  K only has 1 valence electron.  It will leave with only a little effort, leaving behind a positively charged K^+1 atom.

Explanation:  A neutral potassium atom has 19 total electrons.  But only 1 of them is in potassium's valence shell.  Valence shell means the outermost s and p orbitals.  Potasium's electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1.  The 4s orbital is the only orbital in the 4th energy level.  So it has a valency of 1.  This means this electron will be the most likely to leave, since it is the lone electron in the oyutermost energy level (4).  When that electron leaves, the charge on the atom go up by 1.  The atom now has a full valence shell of 3s^2 3p^6, the same as argon, Ar.

What would happen when a substance changes from one phase to another? A. The substance loses or gains heat. B. The average kinetic energy of the substance changes. C. The temperature of the substance changes. D. The molecular motion of the substance changes.

Answers

I think the correct answer from the choices listed above is option B. When a substance changes from one phase to another, the average kinetic energy of the substance changes since the molecules movements are changed as well. Hope this answers the question. Have a nice day.

What is the half-life of a radioisotope if 25.0 grams of an original 200.-gram sample of the isotope remains unchanged after 11.46 days?(1) 2.87 d (3) 11.46 d
(2) 3.82 d (4) 34.38 d

Answers

Answer : The correct option is, (2) 3.82 d

Solution : Given,

As we know that the radioactive decays follow first order kinetics.

So, the expression for rate law for first order kinetics is given by :

k=(2.303)/(t)\log(a)/(a-x)

where,

k = rate constant

t = time taken for decay process  = 11.46 days

a = initial amount of the reactant  = 200 g

a - x = amount left after decay process  = 25 g

Putting values in above equation, we get the value of rate constant.

k=(2.303)/(11.46)\log(200)/(25)=0.1814

Now we have to calculate the half life of a radioisotope.

Formula used :t_(1/2)=(0.693)/(k)

Putting value of 'k' in this formula, we get the half life.

t_(1/2)=(0.693)/(0.1814)=3.820

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1. 200/2=100. 100/2=50. 50/2=25. So that's 3 to get to 25. 

2. 11.46/3=3.82

The answer is (2).

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