The new trend will probably be the technology that's in that certain year. Like the trend in the modern days are all electronics or will have to do with electronics. Hope this helps! :) Pls mark me as the BRAINLIEST!!!

3 0

3 years ago

A 0.245-L flask contains 0.467 mol co2 at 159 °c. Calculate the pressure using the ideal gas law.

lubasha [3.4K]

Answer:

Pressure, P = 67.57 atm

Explanation:

Given the following data;

Volume = 0.245 L
Number of moles = 0.467 moles
Temperature = 159°C
Ideal gas constant, R = 0.08206 L·atm/mol·K

Conversion:

We would convert the value of the temperature in Celsius to Kelvin.

T = 273 + °C

T = 273 + 159

T = 432 Kelvin

To find the pressure of the gas, we would use the ideal gas law;

PV = nRT

Where;

P is the pressure.
V is the volume.
n is the number of moles of substance.
R is the ideal gas constant.
T is the temperature.

Making P the subject of formula, we have;

$P = \frac {nRT}{V}$

Substituting into the formula, we have;

$P = \frac {0.467*0.08206*432}{0.245}$

$P = \frac {16.5551}{0.245}$

Pressure, P = 67.57 atm

4 0

3 years ago

Shale forms from tiny particles of which material?

miskamm [114]

Answer:

a. mica

Explanation:

6 0

3 years ago

In a unimolecular reaction with twice as much starting material as product at equilibrium, what is the value of Keq? Is ΔG o pos

lubasha [3.4K]

Answer : The value of $K_{eq}$ is, 0.5 and $\Delta G$ = positive.

Explanation :

The unimolecular reaction is:

$A\rightarrow B$

In unimolecular reaction, the starting material is 2 times to the product.

$A=2B$ .........(1)

As we know that:

$K_{eq}=\frac{B}{A}$ ...........(2)

Now substitute equation 1 in 2, we get:

$K_{eq}=\frac{\frac{A}{2}}{A}$

$K_{eq}=0.5$

Now we have to calculate the value of $\Delta G^o$ at 298 K.

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = gas constant = 8.314 J/mol.K

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = 0.5

Now put all the given values on the above formula, we get:

$\Delta G^o=-(8.314J/mol.K)\times (298K)\ln (0.5)$

$\Delta G^o=1717.32J/mol$

Thus, the value of $\Delta G^o$ at 298 K is, 1717.32 J/mol

As we know that:

$\Delta G$ = +ve, reaction is non spontaneous

$\Delta G$ = -ve, reaction is spontaneous

$\Delta G$ = 0, reaction is in equilibrium

Thus, the $\Delta G$ = +ve. So, the reaction is non spontaneous.

8 0

3 years ago