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2 years ago

How much energy does it take to melt 2 kg of ice? (Refer to table of constants for water.)

Chemistry

2 answers:

ioda2 years ago

5 0

It would be C

2 kg x 1000 g/kg x 1mol/18.02 x 6.03 kj/mol = 669kj

Ivenika [448]2 years ago

4 0

Your answer should be C :)

You might be interested in

When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

5 0

3 years ago

You crumple a piece of paper in your hands. Not all of the kinetic energy from your hands is used to crumple the paper. Where do

Shkiper50 [21]

Answer:

Explanation:

When you crumble a piece of paper it make a sound plus i just did on apex

3 0

2 years ago

How many moles of water are produced if 5.43 mol PbO2 are consumed?<br><br> ANSWER: 10.9

Arada [10]

Answer: 10.9 mol.

Explanation:

To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.

Using cross multiplication:

1.0 mole of PbO₂ → 2.0 moles of H₂O

5.43 moles of PbO₂ → ??? moles of water

The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.

4 0

3 years ago

Read 2 more answers

A saturated solution can become supersaturated under which of the following conditions

Kaylis [27]

the answer is d. when more solute is added

5 0

3 years ago

Calculate the mole fraction of kbr (molar mass 119.00 g/mol) in a solution made by dissolving 0.30 g kbr in 0.400 l of H2O (d =

julia-pushkina [17]

The mole fraction of KBr in the solution is 0.0001

<h3>How to determine the mole of water</h3>

We'll begin by calculating the mass of the water. This can be obtained as follow:

Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
Density of water = 1 g/mL
Mass of water =?

Density = mass / volume

1 = Mass of water / 400

Croiss multiply

Mass of water = 1 × 400

Mass of water = 400 g

Finally, we shall determine the mole of the water

Mass of water = 400 g
Molar mass of water = 18.02 g/mol
Mole of water = ?

Mole = mass / molar mass

Mole of water = 400 / 18.02

Mole of water = 22.2 moles

<h3>How to de terminethe mole of KBr</h3>

Mass of KBr = 0.3 g
Molar mass of KBr = 119 g/mol
Mole of KBr = ?

Mole = mass / molar mass

Mole of KBr = 0.3 / 119

Mole of KBr = 0.0025 mole

<h3>How to determine the mole fraction of KBr</h3>

Mole of KBr = 0.0025 mole
Mole of water = 22.2 moles
Total mole = 0.0025 + 22.2 = 22.2025 moles
Mole fraction of KBr =?

Mole fraction = mole / total mole

Mole fraction of KBr = 0.0025 / 22.2025

Mole fraction of KBr = 0.0001

Learn more about mole fraction:

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6 0

1 year ago