Which of the following is equal to 1 mole?

60 kg of fuel was completely burnt for an experiment. The amount of heat energy was found to be 180000KJ. Calculate calorific va

FrozenT [24]

Answer:

3000 kJ/kg

Explanation:

The calorific value of a substance is the amount of heat produced per unit mass by the combustion of the substance.

It is given by:

$C=\frac{Q}{m}$

where

Q is the amount of heat released

m is the mass of the fuel

In this problem, we have:

m = 60 kg is the mass of fuel

$Q=180,000 kJ$ is the amount of heat released

Therefore, the calorific value of the fuel is:

$C=\frac{180,000}{60}=3000 kJ/kg$

6 0

3 years ago

10. The statement that no two electrons in the same atom can have the

Fynjy0 [20]

Answer:

10.the pauli exclusive principle statues that, in an atom

or molecules, no two electrons can have same four electronic quantum numbers.as an orbit can contain a maximum of only two electrons.the two electrons must have opposing spins

3 0

2 years ago

Suppose that Daniel has a 3.00 3.00 L bottle that contains a mixture of O 2 O2 , N 2 N2 , and CO 2 CO2 under a total pressure of

Alenkinab [10]

Answer:

Partial pressure O₂ → 2.74 atm

Explanation:

Let's analyse the data given:

Volume → 3L

In the bottle there is a mixture of gases that contains, O₂, N₂ and CO₂.

Total pressure is 4.80 atm

Let's apply the Ideal Gases Law to determine the total moles of the mixture

P . V = n . R. T

4.80 atm . 3L = n . 0.082 . 273K

n = 4.80 atm . 3L / 0.082 . 273K → 0.643 moles

We apply the concept of mole fraction:

Mole fraction of a gas X = moles of gas X / Total moles

Mole fraction of a gas X = Partial pressure X / Total pressure

In a mixture, sum of mole fraction of each gas = 1

We determine mole fraction of N₂ → 0.230 / 0.643 = 0.357

We determine mole fraction of CO₂ → 0.350 atm / 4.80 atm = 0.0729

1 - mole fraction N₂ - mole fraction CO₂ = mole fraction O₂

1 - 0.357 - 0.0729 = 0.5701 → mole fraction O₂

We replace in the formula: Mole fraction O₂ = Partial pressure O₂ / 4.80 atm

0.5701 . 4.80 atm = Partial pressure O₂ → 2.74 atm

5 0

3 years ago

Read 2 more answers

on a trip to italy, Cheyne rented a car to tour the country. As he turned onto the highway, he noticed that the speed limit was

rjkz [21]

The answer is 59mph.

8 0

3 years ago

Read 2 more answers

Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin

Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0

3 years ago