(NH4)2SO4 + Ba(NO3)2 reaction and what type of reaction is it

which has a higher entropy for the reaction: 2nh3(g)→n2(g) 3h2(g) which has a higher entropy for the reaction: reactants product

krok68 [10]

The one that has a higher entropy for the reaction is products.

<h3>What is entropy?</h3>

Entropy is a measureable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty. It is the measurement of the amount of thermal energy per unit of temperature in a system that cannot be used for productive work. It is a measure of a system's molecular disorder or unpredictability since work is produced by organized molecular motion.

It should bm be noted that the entropy of gas is more than entropy of aqueous which is more than the entropy of liquid and the entropy of solid.

On the product side there are more gas than the reactant side. Therefore, product has more entropy.

Learn more about entropy on:

brainly.com/question/6364271

#SPJ1

3 0

1 year ago

What is the balanced equation for the reaction of a solution of sodium sulfate is mixed with strontium chloride?

natta225 [31]

Answer:

NaSO⁴(ads) ,ganadicate

6 0

2 years ago

Hello,

nikitadnepr [17]

Answer:

$a)\ 2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3\\b)\ Decomposition\ Reaction$

Explanation:

Ferrous Sulphate $(FeSO4)$ is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide $(SO_3)$ , Sulphur Dioxide $(SO_2)$ as well as Ferric Oxide $(Fe_2O_3)$ .

We can hence, frame a skeletal equation of this reaction and try to balance it.

Hence,

$FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

Now,

a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of steps:

1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside $FeSO_4$ .

2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:

Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :

$2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds decompose into smaller elements and compounds. Here, as Ferrous Sulphate decomposes into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is Decomposition Reaction.

7 0

2 years ago

Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose

Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution = 101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0

2 years ago

Please answer these about Charles law

NNADVOKAT [17]

Answer:

1. V2.

2. 299K.

3. 451K

4. 0.25 x 451 = V2 x 299

Explanation:

1. The data obtained from the question include:

Initial volume (V1) = 0.25mL

Initial temperature (T1) = 26°C

Final temperature (T2) = 178°C

Final volume (V2) =.?

2. Conversion from celsius to Kelvin temperature.

T(K) = T (°C) + 273

Initial temperature (T1) = 26°C

Initial temperature (T1) = 26°C + 273 = 299K

3. Conversion from celsius to Kelvin temperature.

T(K) = T (°C) + 273

Final temperature (T2) = 178°C

Final temperature (T1) = 178°C + 273 = 451K

4. Initial volume (V1) = 0.25mL

Initial temperature (T1) = 299K

Final temperature (T2) = 451K

Final volume (V2) =.?

V1 x T2 = V2 x T1

0.25 x 451 = V2 x 299

6 0

3 years ago