Answer:

Explanation:
![\rm MX(s) $\, \rightleftharpoons \,$ M$^{+}$(aq) + $^{-}$(aq); $K_{\text{sp}}$ = [M$^{+}$][X$^{-}$]\\\\\text{$K_{\text{sp}}$ gives us information on}\\\\\boxed{\textbf{ the equilibrium between the solid and its ions in solution}}](https://tex.z-dn.net/?f=%5Crm%20MX%28s%29%20%24%5C%2C%20%5Crightleftharpoons%20%5C%2C%24%20M%24%5E%7B%2B%7D%24%28aq%29%20%2B%20%24%5E%7B-%7D%24%28aq%29%3B%20%24K_%7B%5Ctext%7Bsp%7D%7D%24%20%3D%20%5BM%24%5E%7B%2B%7D%24%5D%5BX%24%5E%7B-%7D%24%5D%5C%5C%5C%5C%5Ctext%7B%24K_%7B%5Ctext%7Bsp%7D%7D%24%20gives%20us%20information%20on%7D%5C%5C%5C%5C%5Cboxed%7B%5Ctextbf%7B%20the%20equilibrium%20between%20the%20solid%20and%20its%20ions%20in%20solution%7D%7D)
It tells us nothing about the amount of precipitate that will form or the temperature at which the equilibrium occurs.
Answer:
1) The reaction rate is double with respect to that reactant
Explanation:
Hello,
By considering the rate law:

If we double the reactant A concentration, by definition, the rate will be doubled as well since the
power is one (order 1), this could be proved just by checking it out in the rate law.
Best regards.
The number of moles that are contained in the given mass of propane (
is 1.7143 moles.
<u>Given the following data:</u>
- Mass of propane = 75.6 grams.
<u>Scientific data:</u>
- The molar mass of propane = 44.1 g/mol.
To calculate the number of moles that are contained in the given mass of propane (
):
<h3>How to calculate the moles of a compound.</h3>
In this exercise, you're required to determine the number of moles of propane that are contained in the given sample:
Mathematically, the number of moles contained in a chemical compound is given by this formula:

Substituting the given parameters into the formula, we have;

Number of moles = 1.7143 moles.
Read more on number of moles here: brainly.com/question/3173452
Answer 1) Option D : Discrete
Explanation : As in the example the scientist studies about the effects of growing human populations on the biodiversity which is found in a region, and where each region was selected had a different population density of humans from 1 to 10 million per 10 square miles. Then in each region the number of different species that can be found was recorded. So here the numerical data was collected for different regions. So, we can conclude it as discrete because when the variable takes on a countable number of values it is called as discrete.
Answer 2) Option D : The entertainment industry
Explanation : When people had enough time and money with them there was no need of creating the communications industry as it was a need not a luxurious thing. With the creation of Internet industry it is somewhat irrelevant. They had enough of time and money so building a labor industry seems to be a lame choice. So, the option of entertainment industry suits the best.
Answer:
Explanation:
At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).
Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.
Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).
A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.
<u>a. C₂H₄:</u>
- C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)
Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.
The following analysis just shows that the other options are not right.
<u>b. C₂H₂:</u>
- 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)
The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.
<u>с. С₃Н₈</u>
- C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)
The mole ratio is 1 mol C₃H₈ : 5 mol O₂
<u>d. C₂H₆</u>
- 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)
The mole ratio is 2 mol C₂H₆ : 7 mol O₂