A student would like to determine how heating a liquid changes its volume. The student hypothesizes that the liquid will increas
1 answer:
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Answer:
6.82 g H₂S
General Formulas and Concepts:
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
0.200 mol H₂S
<u>Step 2: Identify Conversions</u>
Molar Mass of H - 1.01 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
6.818 g H₂S ≈ 6.82 g H₂S
Answer:
Explanation:
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In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:
Thus, in terms of masses, specific heats and temperatures we can write:
Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:
Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:
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