Which idea comes engliment and is reflected in the constitution

4.0 L sample of methane gas is collected at 30.0c, what is the volume of the sample at -8c?

Vedmedyk [2.9K]

This particular law is a gas law, called Charle's Law. The formula is:

V1 V2
---- = ----
T1 T2

So we know our original volume is 4.0L, so we would plug that into our V1. We know T1 is the 30 degrees, since it relates to our original volume. However, we need to convert it to kelvin. We do this simply by adding 273 degrees to the 30 degrees, since 273 is the constant for kelvin.

We do not know our second volume, however we know out T2. It is -8 degrees, and don't forget to convert it to Kelvin.

So, when we plug all of these numbers into the equation, we are left with V2 to find. To do this we cross multiply (V1 x T2) and then divide by T1. That leaves us with the number for V2. Don't forget to round to the least # of sig figs! And you can divide V1 by T1, and then divide V2 by T2, to ensure your answers are the same, since they are directly porportional and need to be equal to each other.

Hope I could help!

8 0

4 years ago

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 K is given below. 2 H2(g) + O2(g) → 2

DENIUS [597]

Answer:

(a) ΔG° = -474 kJ/mol; E° = 1.23 V

(b) ΔH° negative; ΔS° negative

(c) Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.

Explanation:

Let's consider the following reaction.

2 H₂(g) + O₂(g) → 2 H₂O(l)

with an equilibrium constant K = 1.34 × 10⁸³

(a) Calculate E° and ΔG° at 298 K for the fuel-cell reaction.

We can calculate the standard Gibbs free energy (ΔG°) using the following expression:

ΔG° = - R × T × lnK

ΔG° = - 8.314 × 10⁻³ kJ . mol⁻¹.K⁻¹ × 298 K × ln 1.34 × 10⁸³ = -474 kJ/mol

To calculate the standard cell potential (E°) we need to write oxidation and reduction half-reactions.

Oxidation: 2 H₂ ⇒ 4 H⁺ + 4 e⁻

Reduction: O₂ + 4 e⁻ ⇒ 2 O²⁻

The moles of electrons (n) involved are 4.

We can calculate E° using the following expression:

$E\°=\frac{0.0591V}{n} .logK\\E\°=\frac{0.0591V}{4} .log1.34 \times 10^{83}=1.23V$

(b) Predict the signs of ΔH° and ΔS° for the fuel-cell reaction. ΔH°: positive negative ΔS°: positive negative

The standard Gibbs free energy is related to the standard enthalpy (ΔH°) and standard entropy (ΔS°) through the following expression:

ΔG° = ΔH° - T.ΔS°

Usually, the major contribution to ΔG° is ΔH°. So, if ΔG° is negative (exergonic), ΔH° is expected to be negative (exothermic).

The entropy is related to the number of moles of gases. There are 3 gaseous moles in the reactants and 0 in the products, so the final state is predicted to be more ordered than the initial state, resulting in a negative ΔS°.

(c) As temperature increases, does the maximum amount of work obtained from the fuel-cell reaction increase, decrease, or remain the same?

The maximum amount of work obtained depends on the standard Gibbs free energy.

wmax = ΔG° = ΔH° - T.ΔS°

Since ΔS is negative, as T increases, ΔG becomes more positive. Therefore, the maximum work obtained will decrease as T increases.

5 0

4 years ago

Which of the following is correct concerning Hess's law?

Margaret [11]

I believe the correct answer would be that the change in enthalpy can be found by adding the enthalpies of the individual thermochemical reactions of a chemical reaction. In Hess' Law, enthalpy is independent of the mechanism of the reaction. The enthalpy should be the sum of all the changes in the reaction.

3 0

3 years ago

Read 2 more answers

How is the universe like rising raisin bread dough

Free_Kalibri [48]

what kind of question is this and idk

6 0

3 years ago

Read 2 more answers

Please help (show work if needed)

vlada-n [284]

Basically, this figure represents scattering of alpha particles by the atoms of a gold foil.

Here, A represents that a few alpha particles are deflected through large angle. It shows that the nucleus of the atom is positively charged which repels the positively charged alpha particles and deflect them from their original path.
B represents that most alpha particles pass straight through the gold foil without any deflection from their original path. The reason is that there is a lot of empty space in the atom.
C represents that some alpha particles are deflected through small angles. The reason is same as that of A.

Hope you could understand.

If you have any query, feel free to ask.

3 0

3 years ago