A 50-gram sample of a radioactive nuclide is left to decay for 10 hours. At the end of 10 hours, only 25 grams of the sample is
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<h3>Answer:</h3>
1.39 × 10²³ particles CuCr₂O₇
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right<u> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
[Given] 64.5 g CuCr₂O₇
[Solve] particles CuCr₂O₇
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Cu - 63.55 g/mol
[PT] Molar Mass of Cr - 52.00 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of CuCr₂O₇ - 63.55 + 2(52.00) + 7(16.00) = 279.55 g/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>
1.38944 × 10²³ particles CuCr₂O₇ ≈ 1.39 × 10²³ particles CuCr₂O₇
Answer:
Explanation:
Hello,
In this case, we find the following states:
a. Liquid salt water at 28.0 °C.
b. Liquid salt water at 102.5 °C.
c. Vapor salt water at 102.5 °C.
The first process (1) is to heat the liquid water from 28.0 °C to 102.5 °C and the second one (2) to vaporize the liquid salt water. In such a way, each process has an amount of energy that when added, yields the total energy for the process as shown below:
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