<h2><u>QUE</u><u>STION</u></h2>
It refers to a charged particle or atom.
<h2><u>CHOI</u><u>CES</u></h2>
<u>A.</u><u> </u><u>molecule</u>
B. bromine
C. potassium
D. sulfur
<h2><u>ANSWER</u></h2>
<h3><u>C</u><u>.</u><u> </u><u>pottasium</u></h3>
Answer:
The moles of sucrose that are available for this reaction is 0.0292 moles
Explanation:
Combustion is an specifyc reaction where the reactants react with O₂ in order to produce CO₂ and H₂O
This combustion is: C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
We have to conver the mass to moles, to find out the limiting reactant
10 g . 1 mol / 342 g = 0.0292 moles of sucrose
8 g . 1mol / 32g = 0.250 moles of O₂
The moles of sucrose that are available for this reaction is 0.0292 moles
Before we start to work with the equation we must find the limiting reactant. When you find it, you can do all the calculations.
Answer:
I has 2 double carbon carbon bonds
You can create two conversion factors from one equivalence statement.
For example, consider the equivalence statement
1 kg ≡ 1000 g
If you divide <em>both sides</em> by 1 kg, you get the conversion factor:
1 = <em>1000 g/1 kg</em>.
If you divide <em>both sides</em> by 1000 g, you get the conversion factor:
<em>1 kg/1000 g</em> = 1.
Both of these are conversion factors because <em>they both equal one</em>, and multiplying a measurement by one does not change its value.