Answer:
red supergiants is the answer
Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place
The wheels will be completely used up and it is the limiting reactant in this case.
<h3>What is a limiting reactant?</h3>
The limiting reactant is the reactant that is completely used up in a reaction, and thus determines when the reaction stops.
- 60 breaks will be used for 30 engines and 30 body frame
- 80 wheels will be used for 20 engines and 20 body frame
- 64 headlights will be used for 32 engines and 32 body frame
The wheels will be completely used up and it is the limiting reactant in this case.
Learn more about limiting reactants here: brainly.com/question/14222359
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<h3>
Answer:</h3>
1.2 × 10⁻⁸ mol Pb
<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>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 7.2 × 10¹⁵ atoms Pb
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb
There would be 79 electrons present in each atom of gold. I hope this helps :)
