Ask question

Ask question

All categories

3 years ago

14

100 K = _____ -173°C 0°C 100°C 373°C

2 answers:

Hunter-Best [27]3 years ago

7 0

Answer : The correct option is, $-173^oC$

Explanation :

The conversion used for the temperature from Kelvin to degree Celsius is:

$K=273+^oC$

where,

K = temperature in Kelvin = 100 K

$^oC$ = temperature in degree Celsius = ?

Now put the value in the above conversion, we get the temperature in degree Celsius.

$100=273+^oC$

$^oC=100-273$

$^oC=-173$

Therefore, the temperature in degree Celsius is, $-173^oC$

Dahasolnce [82]3 years ago

6 0

100 k= -173°c...............

You might be interested in

What does this image represent?

Hunter-Best [27]

Explanation:

It represent Alcohol group (—OH)

8 0

3 years ago

How many moles are in 11.82 grams of gold?

kobusy [5.1K]

11.82 x 1 mole / (molar mass)
The molar mass = 197 g/mol

11.82 / 197 = .06

5 0

3 years ago

Is this statement true or false? Bases will turn litmus paper red.<br> a. True<br> b. False

abruzzese [7]

False, it will turn the litmus paper blue

3 0

3 years ago

In the compound P2O5, the ratio of phosphorous to oxygen is

pickupchik [31]

The ratio of phosphorous to oxygen is 2:5

<u>Explanation</u>:

In a compound P2O5, the ratio of phosphorous to oxygen atoms is 2:5. Because there are two phosphorous atoms and five oxygen atoms. Both combined are called a P2O5.
The elements of phosphorous and oxygen are added to P2O5. So P2O5 is called as phosphorous pentoxide. The element phosphorous will be added to oxygen and it will form a chemical reaction.
This includes a change in the energy of P2O5. So the new energy will be produced.

5 0

4 years ago

Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit

kati45 [8]

<u>Answer:</u> The $\Delta G$ of the reaction at given temperature is -12.964 kJ/mol.

<u>Explanation:</u>

For the given chemical reaction:

$CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)$

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}$

We are given:

$p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm$

Putting values in above equation, we get:

$K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = $5.34\times 10^{-3}$

Putting values in above equation, we get:

$\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol$

Hence, the $\Delta G$ of the reaction at given temperature is -12.964 kJ/mol.

5 0

4 years ago