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yaroslaw [1]
3 years ago
13

When 25.0 grams of solid potassium hydroxide (KOH, molar mass = 56.1 g/mol) is dissolved in 100.0 grams of water, the solution i

ncreases in temperature by 61.5ºC. What is the molar heat of solution (ΔH) for potassium hydroxide in kJ/mol?
Remember that the mass of the solid is not added to the mass of the liquid to determine the mass of the solution. Just use the mass of the water.

q=CpmΔT and Cp for H2O = 4.18 J/gºC

Select one:
a. +57.7 kJ/mol
b. -1.03 kJ/g
c. -57.7 kJ/mol
d. 57700 J/mol
Chemistry
1 answer:
Maru [420]3 years ago
4 0

Answer:

They gave you the equation; Cp=,

just plug everything in! You’ve seen this; I have long ago, but we had different units. Sorry, but it’s right there! Go get it!

Explanation:

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nekit [7.7K]
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4 0
3 years ago
Please HELP will give Brainliest!!!!!
dedylja [7]
4 mol / 205g H2O = 4/.205 = 19.5 mol/kg boiling point = 100 + 19.5 • 0.51 = 109 ºC
7 0
3 years ago
Why is it important to pasteurize certain liquids?
galben [10]

Answer:

Pasteurization involves heating liquids at high temperatures for short amounts of time, it can make it better if you wanted hot cocoa but bad for milk

Explanation:

4 0
3 years ago
Read 2 more answers
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
Which sphere is dependent on all the other spheres in order to exist?
egoroff_w [7]

Answer:

biosphere

Explanation:

Living things need water (hydrosphere), chemicals from the atmosphere, and nutrients gained by eating things in the biosphere.

6 0
3 years ago
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