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grigory [225]
3 years ago
7

Which list of the phases of H2O is arranged in order of increasing entropy?

Chemistry
1 answer:
Yanka [14]3 years ago
4 0
The answer is (2). The entropy is measured the disorder degree. So the increasing of entropy of the phases are solid, liquid and gas. So the answer is ice, liquid water and steam.
You might be interested in
PLEASE HELP!!
grandymaker [24]

Answer:

1. 136 °C.

2. 0.21 atm.

Explanation:

1. Determination of the new temperature in °C.

Initial volume (V1) = 1.35L

Final volume (V2) = 1.95L

Initial temperature (T1) = 283 K

Final temperature (T2) =...?

Using the Charles' law equation, the new temperature of the gas can be obtained as follow:

V1 /T1 = V2 /T2

1.35/283 = 1.95/T2

Cross multiply

1.35 × T2 = 283 × 1.95

1.35 × T2 = 551.85

Divide both side by 1.35

T2 = 551.85/1.35

T2 = 408.8 ≈ 409 K

Finally, we shall convert 409 K to °C. This can be obtained as follow:

T (°C) = T(K) – 273

T(K) = 409 K

T (°C) = 409 – 273

T (°C) = 136 °C

Therefore, the new temperature of the gas is 136 °C.

2. Determination of the new pressure.

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1.67 L

Final pressure (P2) =.?

Next, we shall convert 1.67 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

1.67 L = 1.67 L × 1000 mL / 1 L

1.67 L = 1670 mL

Therefore, 1.67 L is equivalent to 1670 mL.

Finally, we shall determine the new pressure of the gas as follow:

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1670 mL

Final pressure (P2) =.?

P1V1 = P2V2

1.34 × 267 = P2 × 1670

357.78 = P2 × 1670

Divide both side by 1670.

P2 = 357.78 / 1670

P2 = 0.21 atm.

Therefore, the new pressure of the gas is 0.21 atm.

3 0
3 years ago
A quantity of water is heated from 25.0°C to 36.4°C by absorbing 325 J of heat energy. If the specific heat of water is 4.18 J /
Arlecino [84]

Answer:

6,8 g

Explanation:

c = 4.18 J/(g * °C) = 4180 J / (kg * °C)

t_{1} = 25 °C

t_{2} = 36,4 °C

Q = 325 J

The formula is: Q = c * m * (t_{2} - t_{1})

m = \frac{Q}{c * (t_{2} - t_{1} )}

Calculating:

m = 325 / 4180 * (36,4 - 25) ≈ 0,0068 kg = 6,8 g

6 0
3 years ago
What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at −30.0 °C?
Wewaii [24]

Answer:

The total amount of heat released  is  68.7 kJ

Explanation:

Given that:

mass of water = 94.0 g

moles of water = 94 / 18.02 = 5.216

80⁰C   ------>  0⁰C  -------->    -30⁰C

Q1 = m Cp dT

      = 94 x 4.184 x (0 - 80)

     = -31463.68 J

     = -31.43 kJ

Q2 = 6.01 x 10^3 x 5.216

    = - 31348.16 J

   = -31.35 kJ

Q3 = - 94 x 2.09 x 30

    = - 5893.8 J

   = -5.894 kJ

Total heat = Q1 + Q2 + Q3  = -31.43 kJ  + (-31.35 kJ  ) + (-5.894 kJ )  = -68.7 kJ

Total heat released = -68.7 kJ

Note that the "negative sign" simply indicates heat released, therefore no need to put it in the answer.

6 0
3 years ago
Where are stars that are in their giant or super giant stages located on the Hertzsprung-Russell diagram?
maxonik [38]
I’m pretty sure it’s A) upper right!
3 0
3 years ago
6. What is the oxidation number for the atom indicated in the following compounds.
Hatshy [7]

Answer:

a. +6;

b. +5;

c. +3.

Explanation:

Start with elements with well-known oxidation states.

The oxidation state on oxygen O in compounds is mostly -2. Common exceptions include:

  • -1 in peroxides and
  • positive when oxygen bonds to fluorine.

The oxidation state on group 1 metals (Li, Na, K, etc.) in compounds is mostly +1.

The oxidation state on group 2 metals (Be, Mg, Ca, etc.) in compounds is mostly +2.

Barium Ba is a group 2 metal. The oxidation state on Ba in the compound BaSO₄ is expected to be +2.

The oxidation state on hydrogen H in compounds is mostly +1. The oxidation state on H might be negative when it is bonded to metals.  

The oxidation state on halogens (F, Cl, Br, etc.) is mostly -1. The oxidation state may vary when the halogen is bonded to oxygen or another halogen element.

Compounds are neutral. The oxidation state on all atoms in a compound shall add up to 0. Both BaSO₄ and HClO₂ are neutral.

<h3>BaSO₄</h3>

Oxidation states:

  • Ba: +2;
  • The oxidation state on sulfur S is to be determined;
  • O: -2.

Let the oxidation state on S be x.

2 + x + 4 × (-2) = 0;

x = 6.

Hence, the oxidation state on S in BaSO₄ is +6.

<h3>HClO₂</h3>

Oxidation states:

  • H: +1;
  • Cl here is bonded to oxygen. The oxidation state on chlorine Cl is to be determined;
  • O: -2.

Let the oxidation state on Cl be x.

<em>Refer to the equation in BaSO₄ as an example. Try setting up the equation on your own. </em>

x = 3.

Hence, the oxidation state on Cl is +3.

<h3>PO₄³⁻</h3>

Ions carry charge. Oxidation states on atoms in an ion shall add up to the charge of the ion. The superscript of an ion shows its charge. The superscript 3- in the phosphate ion shows that the ion carries a charge of -3.

Oxidation states:

  • The oxidation state on P is to be found;
  • O: -2.

Let the oxidation state on P be x.

x + 4 × (-2) = -3;

x = 5.

Hence, the oxidation state on P is +5.

4 0
3 years ago
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