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myrzilka [38]
4 years ago
5

Quinine, an antimalarial drug, is 8.63% nitrogen. There are two nitrogen atoms per molecule. What is the molecular weight of qui

nine? (Atomic weight: N = 14.01).
Chemistry
1 answer:
Furkat [3]4 years ago
6 0

Answer:

324.18 g/mol

Explanation:

Let the molecular mass of the antimalarial drug, Quinine is x g/mol

According to question,

Nitrogen present in the drug is 8.63% of x

So, mass of nitrogen = \frac {8.63}{100}\times x

Also, according to the question,

2 atoms are present in 1 molecule of the drug.

Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)

So, mass of nitrogen = 14.01×2 = 28.02

These 2 must be equal so,

\frac {8.63}{100}\times x=28.02

solving for x, we get:

<u>x = 324.18 g/mol</u>

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A hypothetical covalent molecule, x–y, has a dipole moment of 1.44 d and a bond length of 163 pm. calculate the partial charge o
Aneli [31]
As we know,
                                     1 D  =  3.34 × 10⁻³⁰ C.m
So,
                                     1.44 D  =  ?
Solving for 1.44 D,
                                     =  (3.34 × 10⁻³⁰ C.m × 1.44 D) ÷ 1 D
                    
                         1.44 D  =  4.80 × 10⁻³⁰ C.m

Dipole Moment 
is given as,
 
                         Dipole Moment  =  q  ×  r    
Solving for q,
                         q  =  Dipole Moment / r    ------ (1)
Where,
                         Dipole Moment  =  4.80 × 10⁻³⁰ C.m

                         r  =  163 pm  =  1.63 × 10⁻¹⁰ m

Putting values in eq. 1,

                            q  =  4.80 × 10⁻³⁰ C.m / 1.63 × 10⁻¹⁰ m

                            q  =  2.94 × 10⁻²⁰ C

As,
                            1.602 × 10⁻¹⁹ C  =  1 e⁻
So,
                             2.94 × 10⁻²⁰ C  =  X e⁻

Solving for X,

                            X  =  (2.94 × 10⁻²⁰ C × 1 e⁻) ÷ 1.602 × 10⁻¹⁹ C

                                = 0.183 e⁻

Result:
           
So one element is containing + 0.183 e⁻ while the other element is containing - 0.183 e⁻.
4 0
3 years ago
At 20 ∘C the vapor pressure of benzene C6H6 is 75 torr and that of toluene C7H8 is 22 torr. Assume that benzene and toluene fo
Olegator [25]

Answer:

a) Xbenzene = 0.283

b) Xtoluene = 0.717

Explanation:

At T = 20°C:

⇒ vapor pressure of benzene (P*b) = 75 torr

⇒ vapor pressure toluene (P*t) = 22 torr

Raoult's law:

  • Pi = Xi.P*i

∴ Pi: partial pressure of i

∴ Xi: mole fraction

∴ P*i: vapor pressure at T

a) solution: benzene (b) + toluene (t)

∴ Psln = 37 torr;   at T=20°C

⇒ Psln = Pb + Pt

∴ Pb = (Xb)*(P*b)

∴ Pt = (Xt)*(P*t)

∴ Xb + Xt = 1

⇒ Psln = 37 torr = (Xb)(75 torr) + (1 - Xb)(22 torr)

⇒ 37 torr - 22 torr = (75 torr)Xb - (22 torr)Xb

⇒ 15 torr = 53 torrXb

⇒ Xb = 15 torr / 53 torr

⇒ Xb = 0.283

b) Xb + Xt = 1

⇒ Xt = 1 - Xb

⇒ Xt = 1 - 0.283

⇒ Xt = 0.717

6 0
3 years ago
For a particular reaction, ΔH=−111.4 kJ/mol and ΔS=−25.0 J/(mol·K).
PtichkaEL [24]
That’s the answer ^^^^
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3 years ago
Explain why the amino group of p-aminobenzoic acid does not participate in the reaction
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This is because amino group of p-aminobenzoic acid is an aniline and is less electrophilic than an alkyl amine.

<h3>What is an Aniline?</h3>

This is an aromatic amine which consists of a phenyl group attached to an amino group.

The amino group of p-aminobenzoic acid being an aniline makes it less electrophilic which is why an alkyl amine participates in the reaction instead.

Read more about Aniline here brainly.com/question/9982058

5 0
2 years ago
A 150.0 mL sample of a 1.50 M solution of CuSO4 is mixed with a 150.0 mL sample of 3.00 M KOH in a coffee cup calorimeter. The t
svp [43]
Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225 
<span>mols KOH = 3.00 x 0.150 = 0.450 </span>
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<span>CuSO4 + 2KOH = Cu(OH)2 + 2H2O </span>
<span>q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T </span>
<span>q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) </span>
<span>dHrxn in J/mol= q/0.225 mol CuSO4 </span>
<span>Then convert to kJ/mol


</span>
5 0
3 years ago
Read 2 more answers
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