3. The nucleus:
1 answer:
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Answer:
24.9 L Ar
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<u>Aqueous Solutions</u>
- States of Matter
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 40.0 g Ar
[Solve] L Ar
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Ar - 39.95 g/mol
[STP] 22.4 L = 1 mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
24.9235 L Ar ≈ 24.9 L Ar
Taking into account the definition of calorimetry ans sensible heat, you need a heat of 38.72 kJ to raise the temperature of 400 g of ethanol from 20 °C to 60 °C.
<h3>What is calorimetry and sensible heat</h3>Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where
- Q is the heat exchanged by a body of mass m.
- c is the specific heat substance c.
- ΔT is the temperature variation.
In this case, you know:
- Q= ?
- c= 2.42
- m= 400 g
- ΔT= Tfinal - Tinitial= 60 °C - 20 °C= 40 °C
Replacing in the definition of sensible hear:
Q = 2.42 × 400 g× 40 °C
Solving:
<u><em>Q= 38,720 J= 38.72 kJ</em></u>
Finally, you need a heat of 38.72 kJ to raise the temperature of 400 g of ethanol from 20 °C to 60 °C.
Learn more about sensible heat:
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