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nydimaria [60]
3 years ago
13

2 A(g) + B(g) ---> 2 C(g)Rate = k [A][B]At the beginning of one trial of this reaction, [A] = 3.0 M and [B] = 1.0 M. The obse

rved rate for the formation of C is 0.36 mol L-1 sec-1.What are the units for k, the rate constant?mol-1 L sec-1mol L-1 sec-1L mol-1 sec-1mol-1 L-1 sec
Chemistry
1 answer:
Orlov [11]3 years ago
6 0

Answer:

Lmol⁻¹s⁻¹

Explanation:

The rate law of the given reaction is:-

Rate=k[A][B]

Wherem, k is the rate constant.

Given that:-

Rate = 0.36 mol/Lsec = 0.36 M/sec

[A] = 3.0 M

[B] = 1.0 M

Thus,

Applying in the equation as:-

0.36 M/sec =k × 3.0 M× 1.0 M

k = 0.12 (Ms)⁻¹ = 0.12 Lmol⁻¹s⁻¹

<u>The units of k = Lmol⁻¹s⁻¹</u>

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Explain why need modifications about temperature and keep pressure atom
Yuri [45]

Answer:

The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. ... This means that they have more collisions with each other and the sides of the container and hence the pressure is increased.

7 0
3 years ago
What is the molar mass of just the O in CO2?
Vlad1618 [11]

Answer:

16 g/mol

Explanation:

In CO2, it means we have 1 mole of carbon and 2 moles of oxygen.

However, we want to find the molar mass of just a single mole of oxygen.

Now, from tables of values of elements in electronic configuration, the molar mass of oxygen is usually approximately 16 g/mol.

In essence the molar mass is simply the atomic mass in g/mol

8 0
3 years ago
How much heat is gained when a 50.32g piece of aluminum is heated from 9.0°c to 16°c
Rashid [163]

Answer: 317 joules

Explanation:

The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

In this case,

Q = ?

Mass of aluminium = 50.32g

C = 0.90J/g°C

Φ = (Final temperature - Initial temperature)

= 16°C - 9°C = 7°C

Then, Q = MCΦ

Q = 50.32g x 0.90J/g°C x 7°C

Q = 317 joules

Thus, 317 joules of heat is gained.

5 0
3 years ago
Read 2 more answers
Use the molar bond enthalpy data in the table to estimate the value of Δ∘rxn
MakcuM [25]

Answer:

ΔH°rxn = - 433.1 KJ/mol

Explanation:

  • CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)

∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state

∴ ΔH°CCl4(g) = - 138.7 KJ/mol

∴ ΔH°HCl(g) = - 92.3 KJ/mol

∴ ΔH°CH4(g) = - 74.8 KJ/mol

⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)

⇒  ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol

⇒ ΔH°rxn = - 433.1 KJ/mol

4 0
3 years ago
Read 2 more answers
Write the formulas of the three singly chlorinated isomers formed when 2,2-dimethylbutane reacts with Cl2 in the presence of lig
Tems11 [23]

The molecule with same molecular formula but different arrangement of atoms is said to be an isomer.

When 2,2-dimethylbutane reacts with chlorine in the presence of light gives three isomers that is CH_{3}C(CH_{3})_{2}CHClCH_{3} (3-chloro-2,2-dimethylbutane), ClCH_{2}C(CH_{3})_{2}CH_{2}CH_{3} (1-chloro-2,2-dimethylbutane) and ClCH_{2}CH_{2}C(CH_{3})_{2}CH_{3} (1-chloro-3,3-dimethylbutane).

In above case, the molecular formula of all isomers are same i.e.C_{6}H_{13}Cl but chlorine is arranged in different positions of carbon. Thus, results isomers.

The reaction is shown in the image.


3 0
3 years ago
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