What is the value of Kw at 25°C?<br> 1.0 x 10–14 <br> 1.0 x 10

Given the following values for the change in enthalpy (deltaH) and entropy (deltaS), which of the following processes can occur

Shtirlitz [24]

Answer:

Option A and B

Explanation:

(a) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)

Delta G = Delta H – T * DS

Substituting the given values, we get –

Delta G = -84 -298 *(125/1000) = -121.25 KJ/mol

Delta G is negative hence the process is spontaneous and will not violate the second law of thermodynamics

(b) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = -125j mol-2K-1)(-30 cal mol-1 K-1)

Delta G =-84 -298 *(-125/1000) = -46.75 KJ/mol

Delta G is negative hence the process is spontaneous and will not violate the second law of thermodynamics

(c) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)

Delta G = 84 -298 *(125/1000) = +46.75 KJ/mol

Delta G is positive hence the process is non-spontaneous and will violate the second law of thermodynamics

(d) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(-30 cal mol-1 K-1)

Delta G = 84 -298 *(-125/1000) = + 121.25 KJ/mol

Delta G is positive hence the process is non-spontaneous and will violate the second law of thermodynamics

6 0

3 years ago

What is the percentage of hydrogen in c2h4

jarptica [38.1K]

Ethylene- C2H4 = 85.7% Carbon and 14.3% Hydrogen

Find the atomic masses for each element and multiply it by the number of atoms in the compound, then add.

C- 12.0 * 2= 24.0

H- 1.00 * 4= 4.00

-----------------------

28.0

Take the masses for each element and divide it by the total mass. Then change the answer to get the percent.

C 24.0 / 28.0= .857 = 85.7%

H 4.00 / 28.0= .143 = 14.3%

<h3>Ethylene is 85.7% Carbon and 14.3% Hydrogen </h3>

8 0

3 years ago

Given a fixed amount of gas held at constant pressure, calculate the volume it would occupy if a 2.00 L sample were cooled from

aliina [53]

Answer:

1.82 L

Explanation:

We are given the following information;

Initial volume as 2.0 L
Initial temperature as 60.0°C
New volume as 30.0 °C

We are required to determine the new volume;

From Charles's law;

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where, $V_1 and V_2$ are initial and new volume respectively, while $T_1 and T_2$ are initial and new temperatures respectively;

$T_1= 333 K$

$T_2=303K$

$V_1 =2.0L$

Rearranging the formula;

$V_2=\frac{V_1T_2}{T_1}$

$= \frac{(2.0L)(303K)}{333K} \\=1.820 L$

Therefore, the new volume that would be occupied by the gas is 1.82 L

7 0

3 years ago

7. A. How many moles of Calcium hydroxide (CaOH2) are needed to completely react with 14.5 mol of Phosphoric acid (H3PO4)?

Artyom0805 [142]

Answer:

moles of calcium hydroxide= 21.75 mol

a) 43.5 mol

b) 7.25 mol

Explanation:

Please see the attached picture for the full solution.

6 0

3 years ago

How many grams of water will be produced from 50 g hydrogen reacting with 50 g oxygen?

DanielleElmas [232]

We need to first come up with a balanced equation:

$4H+O_{2}$ → $2H_{2}O$

We know that the molar ratio of hydrogen to oxygen to water now is 4:1:2.

Converting the amount of grams given to moles is as follows:

Hydrogen: $\frac{1mole}{1.008g}*50g=49.6mol$

Oxygen: $\frac{1mole}{15.999g}*50g=3.125mol$

We know now that the limiting reactant is oxygen. We can then know that the number of moles of water are produced are double the number of moles of oxygen used due to the ratio that we established at the beginning - 4:1:2.

So we now can use 6.25 moles of water as the amount produced.

Then we convert moles of water to grams:

$\frac{18.015g}{1mole} *6.25mol=112.59g$

Now we know that there are 112.59g of water produced when we start with 50g of hydrogen and 50g of water.

4 0

3 years ago

What is the value of Kw at 25°C? 1.0 x 10–14 1.0 x 10–7 1 7 14