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3 years ago

A system is said to be in a state of dynamic equilibrium when the

Chemistry

1 answer:

timama [110]3 years ago

7 0

Answer:

The correct option is D.

Explanation:

Chemical equilibrium refers to a state in which the concentration of both reactants and products are present in concentration which shows no tendency of changing with time. At this point there is no observable changes in the system. The chemical equilibrium will change to dynamic equilibrium once a reversible reaction occurs. In this case, the ratio of reactants or product changes but there are movements between the chemicals in such a way that there is no net change. At dynamic equilibrium, reactant and products are formed in a rate that keep the concentration of both constant.

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A 0.529-g sample of gas occupies 125 ml at 60. cm of hg and 25°c. what is the molar mass of the gas?

Llana [10]

Let's assume that the gas has ideal gas behavior. 
Then we can use ideal gas formula,
PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.

P = 60 cm Hg = 79993.4 Pa
V = 125 mL = 125 x 10⁻⁶ m³

n = ?

R = 8.314 J mol⁻¹ K⁻¹
T = 25 °C = 298 K

By substitution,
79993.4 Pa x 125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K
n = 4.0359 x 10⁻³ mol

Hence, moles of the gas = 4.0359 x 10⁻³ mol

Moles = mass / molar mass

Mass of the gas = 0.529 g

Molar mass of the gas = mass / number of moles
= 0.529 g / 4.0359 x 10⁻³ mol
 = 131.07 g mol⁻¹

Hence, the molar mass of the given gas is 131.07 g mol⁻¹

4 0

3 years ago

What advice would you give a person who handles hydrogen and oxygen in their workplace?

Murljashka [212]

Regard the principle of utilization of two gas.

Make a consistent control of hardware containing gas.

Make a consistent control of weight diminishing valves giving gas.

No smoking zone.

3 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago

The pH of a 1.0M solution of butanoic acid HC4H7O2 is measured to be 2.41. Calculate the acid dissociation constant Ka of butano

Lubov Fominskaja [6]

Answer:

Ka = 1.52 E-5

Explanation:

CH3-(CH2)2-COOH ↔ CH3(CH2)2COO- + H3O+

⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]

mass balance:

⇒ C CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M

charge balance:

⇒ [H3O+] = [CH3(CH2)2COO-]

⇒ Ka = [H3O+]²/(1 - [H3O+])

∴ pH = 2.41 = - Log [H3O+]

⇒ [H3O+] = 3.89 E-3 M

⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )

⇒ Ka = 1.519 E-5

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3 years ago

Carbon is found in all known forms of life. Which statement best describes why carbon is uniquely suited to forming macromolecul

Helen [10]

C
Carbon and its compound's stability means it is useful in synthesizing organic molecules which are used in the bodies of organisms.

6 0

3 years ago

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