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Ksenya-84 [330]
3 years ago
14

Pls HELP AND EXPLAIN Can Jupiter support LIQUID water?

Chemistry
1 answer:
AURORKA [14]3 years ago
3 0

Answer:

The answer is yes, there is a small amount of water, but it is not ”on” Jupiter. It is in the form of water vapor in the cloud tops.

Explanation:

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When I cut it a gas will irritate my eye. Sometimes when I'm cooking you might see me cry.
Elena-2011 [213]
Lol, onions. Onions produce a chemical irritant called synpropanerhial-S-oxide.
3 0
2 years ago
A silver cube with an edge length of 2.38 cm and a gold cube with an edge length of 2.79 cm are both heated to 81.9 ∘C and place
never [62]

Answer:

The final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

Explanation:

<u>Given data;</u>

edge length of silver, a = 2.38 cm = 0.0238 m

edge length of gold, a = 2.79 cm = 0.0279 m

final temperature of silver, t = 81.9 ° C

final temperature of Gold, t = 81.9 ° C

initial temperature of water, t = 19.6 ° C

volume of water, v =  109.5 mL = 0.0001095 m³

<u>Known data:</u>

density gold 19300 kg/m³

density silver 10490 kg/m³

density water 1000 kg/m³

specific heat gold is 129 J/kgC

specific heat silver is 240 J/kgC

specific heat water is 4200 J/kgC

<u>Calculated data</u>

Apply Pythagoras theorem to determine the side of each cube;

Silver cube;

let L be the side of the silver cube

Taking the cross section of the cube (form a right angled triangle), the edge  length forms the <em>hypotenuse side</em>.

L² + L² = 0.0238²

2L² = 0.0238²

L² = 0.0238² / 2

L² = 0.00028322

L = √0.00028322

L = 0.0168

Volume of cube = L³

Volume of the silver cube = (0.0168)³ = 4.742 x 10⁻⁶ m³

Gold cube;

let L be the side of the gold cube

L² + L² = 0.0279²

2L² = 0.0279²

L² = 0.0279² / 2

L² = 0.0003892

L = √0.0003892

L = 0.0197

Volume of cube = L³

Volume of the silver cube = (0.0197)³ = 7.645 x 10⁻⁶ m³

Mass of silver cube;

density = mass / volume

mass = density x volume

mass of silver cube = 10490 (kg/m³) x 4.742 x 10⁻⁶ (m³) = 0.0497 kg

Mass of Gold cube

mass of gold cube = 19300 (kg/m³) x 7.645 x 10⁻⁶ (m³) = 0.148 kg

Mass of water

mass of water = 1000 (kg/m³) x 0.0001095 (m³) = 0.1095 kg

Let the heat gained by cold water be  Q₁

Let the heat lost  by silver cube = Q₂

Let the heat lost  by gold cube = Q₃

Let the final temperature of water = T

Q₁  = 0.1095 kg x 4200 J/kgC x (T–19.6)

Q₂ = 0.0497 kg x 240 J/kgC x (81.9–T)

Q₃ = 0.148 kg x 129 J/kgC x (81.9–T)

At thermal equilibrium;

Q₁ = Q₂ + Q₃

0.1095 x 4200  (T–19.6)  = 0.0497 x 240 x (81.9–T)  + 0.148 x 129 x (81.9–T)

459.9 (T–19.6) = 11.928 (81.9–T) + 19.092(81.9–T)

459.9T - 9014.04 = 976.9032 - 11.928T + 1563.6348 - 19.092T

459.9T + 11.928T + 19.092T = 9014.04  + 976.9032 + 1563.6348

490.92T = 11554.578

T = 11554.578 / 490.92

T = 23.54 ⁰C

Therefore, the final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

6 0
3 years ago
Balance the equation <br> NaOH + CH3COOH
solong [7]

NaOH + CH3COOH -> CH3COONa + H20

8 0
3 years ago
Calculate the ΔG°' for the reaction with 3 significant figures with no label for the dimension (just the number). fructose-6-pho
seraphim [82]

Answer:

ΔG° = 1747.523

Explanation:

The parameters mentioned are;

Gibbs Free energy ΔG°

Equilibrium constant Kc

Temperature T = 37 + 273 = 310 (upon conversion to kelvin temperature)

The formular relating all three parameters is given as;

ΔG° =  -RTlnKc

Where; R = rate constant = 8.314 J⋅K−1⋅mol−1

Upon solving;

ΔG° = - 8.314 * 310 * ln(1.97)

ΔG° = 1747.523

6 0
3 years ago
( its supposed to be science but whatever )
lana66690 [7]

Answer:

true

Explanation:

because God created darkness and water first on th first day .according to order sources

7 0
3 years ago
Read 2 more answers
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