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3 years ago

Help me please:

Chemistry

1 answer:

vovangra [49]3 years ago

5 0

Answer:

24805.44 J

Explanation:

Step 1:

Data obtained from the question.

Mass (M) = 464g

Initial temperature (T1) = 120°C

Final temperature (T2) = 219°C

Change in temperature (ΔT) = T2 - T1 = 219°C - 120°C = 99°C

Specific heat capacity of lead (C) = 0.129cal/g°C = 4.184 x 0.129 = 0.54J/g°C

Heat (Q) =?

Step 2:

Determination of the heat Q, required the temperature of lead. This is illustrated below

Q = MCΔT

Q = 464 x 0.54 x 99

Q = 24805.44 J

Therefore, 24805.44 J of heat is required to raise the temperature of lead.

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How many grams are in 0.75 moles of SO2

lana66690 [7]

<h3>Answer:</h3>

48 g SO₂

<h3>General Formulas and Concepts:</h3>

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Order of Operations: BPEMDAS

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Division
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<h3>Explanation:</h3>

Step 1: Define

0.75 mol SO₂

Step 2: Identify Conversions

[PT] Molar Mass of S - 32.07 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of SO₂ - 32.07 + 2(16.00) = 64.07 g/mol

Step 3: Convert

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Multiply/Divide: $\displaystyle 48.0525 \ g \ SO_2$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

48.0525 g SO₂ ≈ 48 g SO₂

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