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dusya [7]
2 years ago
8

Example: Sugar (C12H22O11) - Add up the masses of each element. (Look at your periodic table for the masses.)

Chemistry
1 answer:
Leto [7]2 years ago
3 0

Answer:

342.3 g

Explanation:

From the question given above, the following data were obtained:

C = 12.011 g/mol

H = 1.00794 g/mol

O = 15.9994 g/mol

Molar mass of C₁₂H₂₂O₁₁ =?

C = 12 × 12.011 = 144.132 g

H = 22 × 1.00794 = 22.17468 g

O = 11 × 15.9994 = 175.9934 g

Total mass = 144.132 + 22.17468 + 175.9934

Total mass = 342.3 g

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If a temperature increase from 21.0 ∘c to 35.0 ∘c triples the rate constant for a reaction, what is the value of the activation
Airida [17]

Answer:

59.077 kJ/mol.

Explanation:

  • From Arrhenius law: <em>K = Ae(-Ea/RT)</em>

where, K is the rate constant of the reaction.

A is the Arrhenius factor.

Ea is the activation energy.

R is the general gas constant.

T is the temperature.

  • At different temperatures:

<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>

k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.

ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]

∴ ln(3) = 1.859 x 10⁻⁵ Ea

∴ Ea = ln(3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.

4 0
2 years ago
Gallium has two naturally occurring isotopes with the following masses and natural abundances: Isotope Mass ( amu ) Abundance (%
ICE Princess25 [194]
The formula for determination of atomic mass given the mass of isotopes and relative abundance is:

Ar = ∑(mass * abundance) / 100
Ar = (68.92558 * 60.108 + 70.92470 * 39.892) / 100
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The atomic mass of gallium is 69.72306 amu
4 0
3 years ago
An unknown acid solution has PH 3.4. 66% of the acid is ionized. Whats the pka?
juin [17]

Answer:

pKa=3.58

Explanation:

Hello,

In this case, since the pH defines the concentration of hydrogen:

pH=-log([H^+])

[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}

And the percent ionization is:

\% \ ionization=\frac{[H^+]}{[HA]}*100\%

We compute the concentration of the acid, HA:

[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%}  *100\%\\\\

[HA]=6.03x10^{-4}

Thus, the Ka is:

Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\  \\Ka=2.63x10^{-4}

So the pKa is:

pKa=-log(Ka)=-log(2.63x10^{-4})\\\\pKa=3.58

Regards.

5 0
2 years ago
Jenny wants to test the electrical conductivity of two substances dissolved in water. She is preparing the containers for the ex
katen-ka-za [31]

Answer:

Volume of the solutions

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3 0
3 years ago
One isotope of a metallic element has the mass number 67 and 37 neutrons in the nucleus. The cation derived from the isotope has
Anuta_ua [19.1K]
No. Of protons = mass no. - no. Of neutrons
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Symbol - Zn^+2
6 0
3 years ago
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