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Harlamova29_29 [7]
2 years ago
10

Calculate the density of the four solutions. All of the solutions has the volume equal to 25 ml. Red solution has 25.0 g of mass

. Green has 26.5 g of mass. Yellow has 28.2 g of mass and Blue one has 30.0 g of mass.
(mark the one best answer)

A.
Red solution density is1.00 g/ml and the blue solution density is 0.83 g/ml

B.
Green solution density is 1.06 g/ml and blue solution density is 1.20 g/ml

C.
Green solution density is 0.94 m=g/ml and yellow solution density is 0.88 g/ml

D.
Blue solution density is 1.00 g/ml and Red solutio density 1.20 g/ml
Chemistry
1 answer:
Scilla [17]2 years ago
3 0

Answer:

              B. Green solution density is 1.06 g/ml and blue solution density is 1.20 g/ml

Explanation:

Density is given as,

                               D = Mass / Volume

Red Solution,

                               D = 25 g / 25 mL

                               D = 1 g/mL

Green Solution,

                               D = 26.5 g / 25 mL

                               D = 1.06 g/mL

Yellow Solution,

                               D = 28.2 g / 25 mL

                               D = 1.128 g/mL

Blue Solution,

                               D = 30 g / 25 mL

                               D = 1.20 g/mL

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Answer:

d. may melt if heat is transferred to it from hot mantle rock rising up into the crust.

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Hot, solid rock in the Earth's crust is a solid outer region of the Earth, Underneath the Earth Structure lies the mantle which is composed of solid rocks especially granite and basalt. The mantle is very hot and may melt if heat is transferred to it from hot mantle rock rising up into the crust therefore forming a molten magma. It may even be hotter (and not cooler as it was stated in the option A which makes it the wrong answer) than the surface lava because of the pressure of the overlying rock. As such, This effect is seen when a volcanic eruption occurs.

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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
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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

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