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3 years ago

Explain different ways you could increase the reaction rate, and/or decrease the reaction rate

1 answer:

7 0

Ways to increase reaction rate:
*increase the total surface area per volume ratio
*increase the temperature
*increase the pressure
*decrease the size of particles

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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

never [62]

Answer:

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

Explanation:

Given data;

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

Known data:

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

Calculated data

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 hypotenuse side.

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

If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final

mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l) +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

$\tt \dfrac{8.5}{78.11}=0.109$

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

$\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol$

Heat transferred to water :

Q=m.c.ΔT

$\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC$

3 0

3 years ago

Which option is an example of a chemical property?

Alona [7]

Answer:
The ability to decompose
Explanation:
A property is a chemical property when it changes the chemical structure of a substance after a reaction.
Density and color are both physical properties.
Even though melting point may seem like a chemical property, when something melts, only the physical state changes, and the chemical structure does not change, and therefore, is a physical property.
The ability to decompose is a chemical property. When something decomposes, the chemical structure of many molecules change, and therefore, is classified as a chemical property.

8 0

3 years ago

Read 2 more answers

(5.625 + 8.15) x 2.34 + 3.2

dalvyx [7]

The answer is 35.4335

Hope this helped! (Plz mark me brainliest!)

3 0

3 years ago

Read 2 more answers

Hydrofluoric acid, hf, has a ka of 6.8 × 10−4. what are [h3o+], [f−], and [oh−] in 0.710 m hf?

STALIN [3.7K]

Answer:

[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

For a weak acid like HF, the dissociation of HF will be:

HF + H₂O ⇄ H₃O⁺ + F⁻.

[H₃O⁺] = [F⁻].

∵ [H₃O⁺] = √Ka.C,

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.2 x 10⁻²) = 4.55 x 10⁻¹³.

6 0

3 years ago