The name of this compound is Ethyl propyl ether, CH3CH2OCH2CH2CH3 stands for Ethyl propyl ether.
Jellyfish reproduction<span> involves several different stages. In the adult, or medusa, stage of a </span>jellyfish<span>, they can </span>reproduce<span>sexually by releasing sperm and eggs into the water, forming a planula. ... During this stage, which can last for several months or years, asexual </span>reproduction<span> occurs.</span>
Answer:
Mass
Step-by-step explanation:
Usually, you plot the independent variable along the horizontal (x) axis and the dependent variable along the vertical (y) axis.
Marcia's teacher plotted the mass of the sample along the x-axis and volume along the y-axis.
The mass is the independent variable, because that is <em>what the teacher varied</em>.
The volume is the <em>dependent variable</em>, because it <em>depends</em> on the mass.
Sample number is <em>wrong</em>, because it is not a variable.
Substance is <em>wrong</em>, because all samples consist of the same substance.
Density is <em>wrong</em>, because it is constant. It is the slope of the graph.
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.