They are the elements from scandium to copernicium. The elements between group 2 and group 3
Answer:
See Explanation
Explanation:
A decreasing temperature indicates that the dissolution process for the ammonium chloride requires input of energy from surroundings. That is, the process is essentially 2 parts => system (object of interest - NH₄Cl) and the surroundings (everything else - solvent - H₂O). The surroundings (water) solvent is showing a <u>measured</u> decrease in temperature or loss of energy (exothermic to surroundings) which flows into the system (NH₄Cl) and effects dissolution of salt into solution (endothermic to system).
Answer:
Explanation:one is true because if it is positiveit would not be attracting it would be sepperating
Answer:
No because some lights we cannot see because they move to slow or they move to fast for our eye to see. There is just a small little gap compared to what light rays we can actually see. In this picture you can see what I mean.
Explanation:
<h3>
Answer:</h3>
733 g CO₂
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O
[Given] 5.55 mol C₃H₇OH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol C₃H₇OH → 6 CO₂
Molar Mass of C - 12.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
732.767 g CO₂ ≈ 733 g CO₂