Coal like graphite is composed of carbon. Unlike graphite, coal is formed by decayed, once living matter. Is it a mineral?

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

5 0

2 years ago

A nickle is about 2 millimeters thick, or 2/1000 of a meter.how many nanometers is this?

Softa [21]

You know that you can just google it right? It has this converter.

8 0

3 years ago

Why do you think scientists claim to offer only supporting evidence for a theory and not proof?

larisa [96]

Because they're not confident enough to confirm it's indeed true but they will support it for they believe it's correct.

8 0

2 years ago

What will happen if you take a frog out of the food chain?

Setler [38]

<span>there are a number of animals that eat frogs that will die off, then the animals that eat them will go hungry, and their populations will die off as well. This would just keep going up the food chain. Essentially, the loss of the frogs could have deep, reverberating effects that we may not grasp until it’s too late.</span>

4 0

2 years ago

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K

Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

$\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S$

Thus, by combining them, we obtain:

$-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}$

Which is related to the general line equation:

$y=mx+b$

Whereas:

$y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}$

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

8 0

2 years ago