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2 years ago

An organic thiol compound is 38.66% C, 9.73% H, and 51.61% S by mass. What is the empirical formula of this compound?

Chemistry

1 answer:

loris [4]2 years ago

6 0

Answer:

Empirical formula is C₂H₆S

Explanation:

Given data:

Percentage of hydrogen = 9.73%

Percentage of carbon = 38.66%

Percentage of sulfur = 51.61%

Empirical formula = ?

Solution:

Number of gram atoms of H = 9.73 / 1.01 = 9.6

Number of gram atoms of S = 51.61 / 32 = 1.6

Number of gram atoms of C = 38.66 / 12 = 3.2

Atomic ratio:

C : H : S

3.2/1.6 : 9.6/1.6 : 1.6/1.6

2 : 6 : 1

C : H : S = 2 : 6 : 1

Empirical formula is C₂H₆S.

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If honey has a density of 1.36 g/ml what is the mass of 1.25 qt reported in kilograms

yawa3891 [41]

Answer;
1.6 kg.

Solution;
The density is 1.36 g/ml;

The volume is 1.25 qt
However; 1 qt = 946.35 ml

Mass is given by; density × volume;
= 1.25 qts × 946.25 ml/qt × 1.36 g/ml =1609 g
but; 1 kg = 1000 g
Hence the mass = 1609/1000 = 1.609 Kg or 1.61 (sig figs)

7 0

2 years ago

A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

$\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}$

$P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}$

$P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times 35.0} = 0.776969038566$

∴ The final pressure P₂ ≈ 0.78 atm.

4 0

2 years ago

Drinking water may contain a low concentration of lead ion (Pb2 ) due to corrosion of old lead pipes. The EPA has determined tha

Lynna [10]

Answer:

3.861x10⁻⁹ mol Pb⁺²

Explanation:

We can define ppm as mg of Pb²⁺ per liter of water.

We calculate the mass of lead ion in 100 mL of water:

100.0 mL ⇒ 100.0 / 1000 = 0.100 L

0.100 L * 0.0080 ppm = 8x10⁻⁴ mg Pb⁺²

Now we convert mass of lead to moles, using its molar mass:

8x10⁻⁴ mg ⇒ 8x10⁻⁴ / 1000 = 8x10⁻⁷ g

8x10⁻⁷ g Pb²⁺ ÷ 207.2 g/mol = 3.861x10⁻⁹ mol Pb⁺²

4 0

2 years ago

What explains the observation that FADH2 oxidation yields one less ATP than NADH oxidation by the Electron Transport System

Allisa [31]

Answer:

Check explanation

Explanation:

During the Electron Transport System occur in the mitochondrial membrane, oxygen in this reaction is been reduced to water and ATPs are being produced.

The quinone form or the oxidized form of Flavin Adenine Dinucleotide(FAD) is the FADH2. While Nicotinamide adenine dinucleotide is the acronym for NADH. NADH is a good donating substance/agent.

It has been observed that FADH2 produce two(2) ATP while NADH produces three(3) ATP. The reason for this observation is that the production of electron in the FADH2 is at the lower enegy level. Because of this it can not transfer its electron to the first complex .

While;

NADH is at the higher energy level and it can directly transfer its electron to the first complex.

8 0

2 years ago

When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at

STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

$1.22 * 10^{5} = - 8.314* 2400 * ln K$

$\\ 1.22 * 10^{5} = -19953.6 * ln K$

$ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022$

So, the equilibrium constant is 0.0022.

4 0

2 years ago