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irakobra [83]
2 years ago
10

Lysine is used as a topical medicinal skin creme. what is the molecular formula of lysine if the empirical formula is c3h7no, an

d the approximate molar mass is 146 g/mol
Chemistry
1 answer:
spayn [35]2 years ago
5 0
Molecular Formula is calculated using following formula,

Molecular Formula  =  n (Empirical Formula)     ----- (1)

Calculating n;

                      n  =  Molecular Weight / Empirical Formula Weight  --- (2)

Empirical formula weight = (12×3) + (1×7) + (14×1) + (16×1)

Empirical formula weight = (36) + (7) + (14) + (16)

Empirical formula weight = 73

Putting values in Eq. 2,

                                n  =  146 / 73

                                n  =  2

Putting Value of n and Empirical Formula in Eq. 1,

Molecular Formula  =  2 × C₃H₇NO

Molecular Formula  =  C₆H₁₄N₂O₂

Result:
           Molecular Formula of Lysine is C₆H₁₄N₂O₂.
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Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C w
just olya [345]

Answer: 0.036 J/g°C

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = 3.42 Kilojoules

[Convert 3.42 kilojoules to joules

If 1 kilojoule = 1000 joules

3.42 kilojoules = 3.42 x 1000 = 3420J]

Mass = 2.508Kg

[Convert 2.508 kg to grams

If 1 kg = 1000 grams

2.508kg = 2.508 x 1000 = 2508g]

C = ? (let unknown value be Z)

Φ = (Final temperature - Initial temperature)

= 42.061°C - 4.051°C

= 38.01°C

Apply the formula, Q = MCΦ

3420J = 2508g x Z x 38.01°C

3420J = 95329.08g•°C x Z

Z = (3420J / 95329.08g•°C)

Z = 0.03588 J/g°C

Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

Thus, the specific heat of the substance is 0.036 J/g°C

7 0
2 years ago
Why is it advisable to wear long sleeves when students work in a chemistry lab? to provide warmth near the lab refrigerator to p
mote1985 [20]
It  is  advisable  to  wear  long  sleeve  when  when  a  student   is  working   in  a  chemistry  lab so   that   to  protect  arms   from  lab  chemicals. when   someone  enter  the  chemistry  lab  to  wort  or  to  study  should  be  well  prepared  with  appropriate  gears   and  security  measure   to  avoid   injury.
9 0
2 years ago
Read 2 more answers
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
2 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
2 years ago
Which of the following is a valid conversion factor?
VMariaS [17]

Answer:

100 cg/1g

Step-by-step explanation:

    1 cg = 0.01 g     Multiply by 100

100 cg = 1 g

(a) is <em>wrong</em>. The correct conversion factor is 1000 cm³/1 L.

(b) is <em>wrong</em>. The correct conversion factor is 1000 mL/1 L.

(c) is <em>wrong</em>. The correct conversion factor is 1 m/10 dm.

7 0
2 years ago
Read 2 more answers
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