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serious [3.7K]
2 years ago
10

Calculate the amount of heat released to convert 150.0 g of to water to ice at 0ºC.

Chemistry
2 answers:
gladu [14]2 years ago
8 0

Answer:

There correct answer is option 1 (-50100 J)

Explanation:

Step 1: Data given

Mass of water = 150.0 grams

ΔHfus = 334 J/g

Temperature = 0°C

ΔHfus = the amount of energy needed to change phase of 1 mol water to ice

Step 2: Calculate the amount of heat required

Q = m*ΔHfus

⇒Q = the amount of heat reuired = TO BE DETERMINED

⇒m =the mass of water = 150.0 grams

⇒ΔHfus = 334 J/g

Q = 150.0 grams * 334 J/g

Q = 50100 J  of heat required

Step 3: Calculate heat released

Qrequired = -Q released

Qreleased = -50100 J

There correct answer is option 1 (-50100 J)

Svetlanka [38]2 years ago
7 0

Answer:

The answer is -50100 J

Explanation:

I just took the test as I'm in GCA as well (11th grade). My teacher walked us through each formula and gave us multiple practice questions in Jigsaw.

Honestly, all you need to be aware of is the fusion formula which is:

q = m*Hfusion

q = amount of energy

m = mass in grams

Hfusion = 334 J/g (it will not change)

Write down what we're given and what we're looking for:

q = ?

m = 150

Hfusion = 334 J/g

Set up the equation:

q = 150 * 334

Multiply 150 by 334 to get 50100.

You must keep in mind that we're finding how much heat is released which means the energy will be negative.

Add the negative sign to the number to get -50100 Joules

I don't know why that chart is there, it doesn't do anything for this equation. Disregard it for this problem.

Hope I helped! :D

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Answer:

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Explanation:

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<em><u>Using cross multiplication:</u></em>

1.0 mol contains → 6.022 x 10²³ molecules.

??? mol contains → 6.3 x 10²⁴ molecules.

∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(6.3 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 10.46 mol.

<em>∴ The no. of grams of NH₃ present = no. of moles x molar mass</em> = (10.46 mol)(17.0 g/mol) = <em>177.8 g ≅ 178 g.</em>

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How long will it take 10.0 mL of Ne gas to effuse through a porous barrier if it has been observed that 125 minutes are required
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Answer:

88.8 minutes

Explanation:

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Putting Values in eq. 1,

           d  =  (1.03 atm × 146.06 g/mol) ÷ (0.08206 atm.L.mol⁻¹.K⁻¹ × 297 K)

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