CHEMISTRY HIGH SCHOOL

All chemical reactions use reactants in a specific proportion or stoichiometry to form products. The __________ reactant regulates the amount of products produced. A) excess B) limiting C) proportional D) stoichiometric

Answers

Answer 1

Answer:

The correct answer is B) Limiting.

Hope this helped!

Answer 2

Answer:

B: Limiting

Explanation:

The limiting reactant regulates the amount of products produced, because you have the least amount of the limiting reactant.

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Which of the following compounds, when added to water, is most likely to increase the solubility of HgS?A. KNO3
B. Na2SO4
C. KCl

Answers

Answer:

B. $Na_2SO_4$

Explanation:

Hello,

The solubility of HgS is increased by adding $Na_2SO_4$ due to common ion effect which consists on the effect that the higher solubility of $Na_2SO_4$ has over the HgS as the sulfur-based ions modify the solubility product. In such a way, by adding the sulfate, the equilibrium will be shifted rightwards, therefore the solubility is increased.

Best regards.

C. KCl- That is the answer for sure, but I dont want to explain

If the acetylene tank contains 37.0 mol of C2H2 and the oxygen tank contains 81.0 mol of O2, what is the limiting reactant for this reaction?

Answers

Answer: $O_2$ is considered as the limiting reagent for this reaction.

Explanation:

Limiting reagent is the reagent which limits the formation of the product.

Excess reagent is the reagent which is present in excess in a chemical reaction.

For the combustion of acetylene, the reaction follows:

$2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2+2H_2O$

By Stoichiometry,

5 moles of oxygen gas reacts with 2 moles of acetylene.

So, 81 moles of oxygen gas will react with = $(2)/(5)* 81$ = 32.4 moles of acetylene.

As, the required amount of acetylene is less than the given amount. So, it is considered as an excess reagent and oxygen gas is the limiting reagent.

Answer:

Oxygen is the limiting reactant of this reaction.

Explanation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

2 moles of acetylene reacts with 5 moles of oxygen.

Then 37 moles of acetylene will react with :

$(5)/(2)* 37.0 mol=92.5 moles$ of oxygen gas.

But only 81 moles of oxygen gas is present.

Then 81.0 moles of oxygen gas will react with:

$(2)/(5)* 81.0 mol=32.4 moles$ of acetylene.

And still acetylene will remain after.

So, from the above calculation we cans see that oxygen is present in limiting amount which will limit the reaction.Hence, the limiting reactant of this reaction is oxygen.

Two most abundant elements in Earth's core are

Answers

Iron accounts for about 86% of the Earth's core and Silicon is next at only 6%. Iron is by far the most abundant element in the Earth's core.

5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of HCl 0,1M. What's the % of the inicial mixture?The 14,3mL of HCl are necessary to neutralize the KOH.

What are the reactions and what do I need to calculate?

Answers

we know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH

I hope this helps.

A sample of hydrogen gas, has a volume of 8.56 L at a temperature 0oC and pressure of 1.5 atm. Calculate the number of moles of hydrogen present in this gas sample ( Assume that the gas behave ideally)

Answers

According to ideal gas law, there are 0.00565 number of moles of hydrogen present in the gas sample .

What is ideal gas law?

The ideal gas law is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.Ideal gas law was proposed by Benoit Paul Emile Clapeyron in year 1834.It is a thermodynamic equation which has wide applications.

Gases which obey ideal gas law are difficult to exist.

Substituting the values as, P=1.5 atmospheres, V=8.56 L, R=8.314 ,T=0°C=273 K that is , n=PV/RT=1.5×8.56/8.314×273=0.00565 moles.

Thus, there are 0.00565 moles of hydrogen present in the gas sample .

Learn more about ideal gas law,here:

brainly.com/question/28257995

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p = 1.5atm ≈ 1519.88hPa
V = 8.56L
R = 83.1 [hPa*L] / [mol*K]
T = 0°C =273K

pV = nRT |:RT

n = pV / RT

n = [1519.88hPa*8.56L] / [83.1 [hPa*L] / [mol*K] * 273K]
n ≈ 0.57mol

Why it is that evaporation causes cooling,but boiling causes heating?

Answers

in evaporation due to internal heat, kinetic energy of molecules increases and they come to the top and take out that heat with them when they evaporate thus causes cooling
in boiling as heat is given to the molecules so their kinetic energy increases and they start vibrating with great energy and thus causes heating

During evaporation, particles which have a high kinetic energy come out of the body by gaining heat due to which there is a cooling effect on the particle of the body itself. that is why, evaporation causes cooling. For example: sweating

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Part AGiven the following equation, N2O(g) + NO2(g) → 3 NO(g) ΔG°rxn = -23.0 kJ Calculate ΔG°rxn for the following reaction. 18 NO(g) → 6 N2O(g) + 6 NO2(g) Answer Choices To Choose From: A) -3.83 kJ B) -138 kJ C) -23.0 kJ D) 23.0 kJ E) 138 kJ Part B CO2(g)→C(s)+O2(g)ΔH∘rxn= +393.5 kJ (ΔSsys<0, ΔSsurr<0) In addition, predict under what temperatures (all temperatures, low temperatures, or high temperatures), if any, the reaction in part D will be spontaneous A) The reaction is spontaneous at all temperatures. C) The reaction is spontaneous at high temperatures. D) The reaction is nonspontaneous at all temperatures.