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3 years ago

HELP HELP!

Chemistry

2 answers:

ra1l [238]3 years ago

8 0

Evaporation A , as it can

Natalija [7]3 years ago

4 0

Answer: A

Explanation: For the water cycle to work, water has to get from the Earth's surface back up into the skies so it can rain back down and ruin your parade or water your crops or yard. It is the invisible process of evaporation that changes liquid and frozen water into water-vapor gas, which then floats up into the skies to become clouds.

You might be interested in

Find δs∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia:12n2(g) 32h2(g)→nh3(g)

Dima020 [189]

Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).

What is standard entropy?

The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."

Calculation:

Balancing the given reaction following-

1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)

ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]

Here S° = standard entropy of the system

Insert into the aforementioned equation all the typical entropy values found in the literature:

ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]

⇒ΔS° = - 99.4 J/K

Therefore, the standard entropy, ΔS° is -99.4 J/K.

Learn more about standard entropy here:

brainly.com/question/14356933

#SPJ4

5 0

1 year ago

A student places a 100.0°C piece of metal that weighs 85.5 g into 122 mL of 16.0°C water. If the final temperature is 20.2°C, wh

Musya8 [376]

Answer:

The specific heat of the metal is 0.314 J/g°C

Explanation:

Step 1: data given

Temperature of the piece of metal = 100.0 °C

Mass of the metal = 85.5 grams

Volume of water = 122 mL = 122 grams

Temperature of water = 16.0 °C

The final temperature of water = 20.2 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of metal

Heat gained= heat lost

Qgained = - Qlost

Qwater = -Qmetal

Q = m*c* ΔT

m(metal)*c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒m(metal) = mass of metal = 85.5 grams

⇒c(metal) = the specific heat of metal = TO BE DETERMINED

⇒ΔT(metal) = the change of temperature of metal = T2 - T1 = 20.2 - 100 °C = -79.8 °C

⇒m(water) = the mass of water = 122 grams

⇒c(water) = the specific heat of water = 4.184 J/g°C

⇒ΔT(water) = the change of temperature of metal = T2 - T1 = 20.2 - 16.0 °C = 4.2 °C

85.5 *c(metal) * -79.8 = -122 * 4.184 * 4.2

c(metal) * (-6822.9) = -2143.9

c(metal) = 0.314 J/g°C

The specific heat of the metal is 0.314 J/g°C

7 0

3 years ago

Based on the reactivities of the elements involved, which reaction will form products that are more stable than the reactants?

MrRa [10]

Answer:

Exothermic reaction

Explanation:

Since in an exothermic reaction heat is released, the products will be more stable than the reactants.

6 0

2 years ago

Examine the statement.

saveliy_v [14]

My view point is that i disagree and that the rules are completely different

6 0

3 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago