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2 years ago

31. Solve this Gibbs' Free Energy Equation:

Chemistry

1 answer:

Vsevolod [243]2 years ago

8 0

Answer:The standard Gibbs energy change at 300K for the reaction 2A⇔B+C is 2494.

Explanation:

ΔG = free energy at any moment.

ΔGo = standard-state free energy.

R is the ideal gas constant = 8.314 J/mol-K.

T is the absolute temperature (Kelvin)

lnQ is natural logarithm of the reaction quotient

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Please help I need the awnser quick will give brain to first to awnser

gizmo_the_mogwai [7]

Possibly decomposition but not sure

7 0

3 years ago

Read 2 more answers

How much positive charge is in 0.7 kg of lithium? with each atom having 3 protons and 3 electrons. The elemental charge is 1.602

hichkok12 [17]

Explanation:

As a neutral lithium atom contains 3 protons and its elemental charge is given as $1.602 \times 10^{-19} C$ . Hence, we will calculate its number of moles as follows.

Moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.7 \times 1000 g}{7 g/mol}$

= 100 mol

According to mole concept, there are $6.023 \times 10^{23}$ atoms present in 1 mole. So, in 100 mol we will calculate the number of atoms as follows.

No. of atoms = $100 \times 6.023 \times 10^{23}$

= $6.023 \times 10^{25}$ atoms

Since, it is given that charge on 1 atom is as follows.

$3 \times 1.602 \times 10^{-19}C$

= $4.806 \times 10^{-19}C$

Therefore, charge present on $6.023 \times 10^{25}$ atoms will be calculated as follows.

$6.023 \times 10^{25} atoms \times 4.806 \times 10^{-19} C$

$28.95 \times 10^{6}C$

Thus, we can conclude that a positive charge of $28.95 \times 10^{6}C$ is in 0.7 kg of lithium.

3 0

3 years ago

Gases and particles which are put into the air or emitted by various sources are called __________. photochemical smog emissions

Leto [7]

I would say the answer is emissions. These are the particles that are not supposed to be present in air but due to the production of different substances from humans daily activities these substances go with the air we breath. Hope this helped.

7 0

3 years ago

CAN SOMEONE PLEASE HELP?

IgorC [24]

Hey there!:

The 1s, 2s and 2p subshells are completely filled (a maximum of two electrons go into the 1s subshell and a maximum of two electrons go into the 2s subshell. The 2p subshell includes 3 orbitals, with 2 electrons maximum per orbital). The 3s subshell has only one of a maximum of two electrons.

Hope that helps!

4 0

3 years ago

Suppose you are titrating vinegar, which is an acetic acid solution

kotegsom [21]

Answer:

0.373 M

Explanation:

The balanced equation for the reaction is given below:

HC2H3O2 + NaOH —> NaC2H3O2 + H2O

From the balanced equation above, the following were obtained:

Mole ratio of the acid, HC2H3O2 (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, we shall write out the data obtained from the question. This include:

Volume of base, NaOH (Vb) = 32.17 mL

Molarity of base, NaOH (Mb) = 0.116 M

Volume of acid, HC2H3O2 (Va) = 10 mL

Molarity of acid, HC2H3O2 (Ma) =..?

The molarity of the acid solution can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 0.116 x 32.17 = 1

Cross multiply

Ma x 10 = 0.116 x 32.17

Divide both side by 10

Ma = (0.116 x 32.17) /10

Ma = 0.373 M

Therefore, the concentration of the acetic acid is 0.373 M.

8 0

3 years ago