Answer:
1) PV > nRT, because the real volume of the gas would be more than the ideal
volume.
Explanation:
According to the ideal gas equation; PV = nRT. Let us recall that this equation only holds under ideal conditions.
Gases exhibit ideal behavior under high temperature and low pressure. At higher pressure, the real volume of the gas is larger than the ideal volume of the gas.
Thus, at high pressure, PV > nRT, because the real volume of the gas would be more than the ideal volume.
Answer:
1) PV > nRT, because the real volume of the gas would be more than the ideal volume.
Explanation:
just took the test :)
PLEASE HELP I HAVE 19 MINUTES LEFT I WILL MARK BRAINLIEST
How much more acidic is a pH of 4 as compared to a pH of 6.5?
Answer:
316.227766
Explanation:
In calorimetry, energy is measured through heat transfer from one substance to
another. Which of the following is NOT a method of heat transfer?
Answer:
Refraction
Explanation:
Convert 7.8 moles of carbon tetrafluoride into grams.
Answer:
686.43363984 is the answer when 7.8 moles is converted.
When 12.00 moles of potassium chlorate decomposes, how many dm3 of oxygen are produced at 325K and 188 kPa?
2KClO3 →2KCl + 3O2
show work pls
Answer:
258.71 dm³
Explanation:
We'll begin by calculating the number of mole of O₂ produced by the decomposition of 12 moles of KClO₃. This can be obtained as follow:
The balanced equation for the reaction is given below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Therefore, 12 moles of KClO₃ will decompose to produce = (12 × 3)/2 = 18 moles of O₂.
Finally, we shall determine the volume of the O₂. This can be obtained as follow:
Temperature (T) = 325 K
Pressure (P) = 188 KPa
Number of mole (n) = 18 moles
Gas constant (R) = 8.314 KPa.dm³/Kmol
Volume (V) =?
PV = nRT
188 × V = 18 × 8.314 × 325
188 × V = 48636.9
Divide both side by 188
V = 48636.9 / 188
V = 258.71 dm³
Thus, 258.71 dm³ of oxygen were obtained from the reaction.
Calculate how many grams of methane (CH4) are in a sealed 800. mL flask at room temperature (22 °C) and 780. mm of pressure. Show work pls.
"0.0340" mol of CH₄ are in sealed flask.
Methane (CH₄)Methane would also be a greenhouse gas, therefore its existence tends to affect humanity's surface temp as well as weather patterns framework; it is released into the atmosphere from such a wide assortment of life forms as well as biogenic.
According to the question,
Volume, V = 800 mL or, 0.800 L
Temperature, T = 22°C or, 295
Pressure, P = [tex]\frac{780}{760}[/tex] = 1.03 atm
As we know the relation,
The gram of moles will be will be:
→ n = [tex]\frac{PV}{RT}[/tex]
By substituting the values, we get
= [tex]\frac{1.03\times 0.800}{0.08206\times 295}[/tex]
= [tex]\frac{0.824}{242.077}[/tex]
= 0.0340
Thus the response above is correct.
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write half-reactions that show how H2O2 can act as either an oxidizing agent or a reducing agent, and describe where each of these situations occurred in your testing.
Answer:
H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .
Explanation:
When
H2O2 acts as an oxidizing agent
H2O2 + 2e- 2H+---> 2H2O
Reducing agent
H2O2 --> O2 + 2e + 2H+
H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .
explain why hydrogen chloride does not conduct electricity, but a solution of hydrogen chloride and water conduct electricity
What does the cell theory state? Answer F All organisms are composed of a nucleus G All prokaryotes are composed of multiple cells H All prokaryotes are single celled organisms J All organisms are composed of cells
Answer:
(J) All organisms are composed of cells
A student weighs 0.347 g of KHP on a laboratory balance. The KHP was titrated with NaOH and the concentration of the NaOH determined to be 0.110 M. For the second titration, the student correctly diluted 6 M HCl from the reagent shelf using a graduated cylinder to obtain approximately 0.6 M HCl. This solution was titrated with the original NaOH solution. The student calculated the concentration of NaOH from the experiment to be 0.099 M. In which experiment should the student be more confident of the concentration of the NaOH solution
Answer:
Following are the solution to the given question:
Explanation:
Each method through KHP is somewhat more precise since we have weighed that requisite quantity, we exactly know the KHP intensity appropriately. Its initial 6 M HCl concentration was never considered mandatory. They have probably prepared 6 M HCl solution although long ago and could have changed its concentration over even a period.
Calculate the volume of solvent present in a 55.5%
by volume of 10.5 mL alcohol solution.
Answer:
I dont know
Explanation:
good luck
Robert Hooke observed a slice of cork under a microscope. He saw what looked like “pores” or “cells” in it. Hooke can assume that the cork is most likely - Answer F A man made material G A type of living organism H A type of mineral J A petroleum based product
The cork oak tree from which cork is extracted is native to southwest europe and northwest africa. Cork is extracted from cork oak trees without harming the tree. So cork is a type of mineral. The correct option is C.
What is a cork cell?The dead cells without having intercellular spaces are defined as the cork cells. They appear at the periphery of roots and stems when they grow older and increase in girth. They also have a chemical called suberin in their walls.
It is the suberin which makes them impervious to gases and water. The outer protective coat of a tree is called the cork. It is one of the components of bark of the tree. The tissues of bark become old and the secondary mersitem replaces them.
Cork is made up of multiple thick layers and it protects the tree from bacterial or fungal infection.
Thus the correct option is C.
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when rolling a number cube 500 times, how many times you expect to get a 3?
Answer:
[tex]\frac{250}{3}[/tex]
Explanation:
you can expect to get a 3 (theoretically) 1 time every 6 times you roll. A 1/6 chance.
Here's the equation:
[tex]\frac{1}{6} =\frac{x}{500}[/tex]
cross multiply (i think that's what it is called)
500=6x
divide by 6 on both sides:
x=[tex]\frac{250}{3}[/tex] or approx 83 times.
Hope this helps! Lmk if u have more questions <3
The equivalence point of a titration corresponds to which of the following?
O the point where equal volumes of acid and base have been used
O Equivalence point is another term for end point
All of the listed options are true
Equivalence point is defined as the point where the pH indicator changes color
O the point where the acid and base have been added in proper stoichiometric amounts
Answer:
E: the point where the acid and base have been added in proper stoichiometric amounts
Explanation:
Equivalence point in titration is simply the point where the amounts of acid and base used just sufficiently reacts chemically to cause neutralization whereas the endpoint is the point where the indicator of the titration changes colour.
The Equivalence point occurs before the endpoint.
Thus, option E is correct.
Inquiry Extension Consider a reaction that occurs between solid potassium and chlorine gas. If you start with an initial mass of 15.20 g K, and an initial mass of 2.830 g Cl2, calculate which reactant is limiting. Explain how to determine how much more of the limiting reactant would be needed to completely consume the excess reactant. Verify your explanation with an example
The 3.13 g of K would be needed to completely react with the remaining [tex]Cl_2[/tex].
To determine which reactant is limiting, we need to calculate the amount of product that can be formed from each reactant and compare them. The reactant that produces less product is the limiting reactant, since the reaction cannot proceed further once it is consumed.
The balanced chemical equation for the reaction between solid potassium and chlorine gas is:
2 K(s) + [tex]Cl_2[/tex](g) -> 2 KCl(s)
From the equation, we can see that 2 moles of K react with 1 mole of [tex]Cl_2[/tex] to form 2 moles of KCl.
First, we need to convert the masses of K and [tex]Cl_2[/tex] into moles:
moles of K = 15.20 g / 39.10 g/mol = 0.388 mol
moles of [tex]Cl_2[/tex] = 2.830 g / 70.90 g/mol = 0.040 mol
Now, we can use the mole ratio from the balanced equation to calculate the theoretical yield of KCl from each reactant:
Theoretical yield of KCl from K: 0.388 mol K x (2 mol KCl / 2 mol K) = 0.388 mol KCl
Theoretical yield of KCl from [tex]Cl_2[/tex]: 0.040 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) = 0.080 mol KCl
We can see that the theoretical yield of KCl from K is 0.388 mol, while the theoretical yield of KCl from [tex]Cl_2[/tex] is 0.080 mol. Therefore, the limiting reactant is [tex]Cl_2[/tex], since it produces less product.
To determine how much more of the limiting reactant would be needed to completely consume the excess reactant, we can use the stoichiometry of the balanced equation.
We know that 1 mole of [tex]Cl_2[/tex] reacts with 2 moles of K to produce 2 moles of KCl. Therefore, the amount of additional K needed to react with the remaining [tex]Cl_2[/tex] can be calculated as follows:
moles of K needed = 0.040 mol [tex]Cl_2[/tex] x (2 mol K / 1 mol [tex]Cl_2[/tex])
= 0.080 mol K
This means that 0.080 moles of K would be needed to completely consume the remaining [tex]Cl_2[/tex]. We can convert this to a mass by multiplying by the molar mass of K:
mass of K needed = 0.080 mol K x 39.10 g/mol
= 3.13 g K
Therefore, The 3.13 g of K would be needed to completely react with the remaining.
Example verification:
Suppose we had an additional 0.50 g of [tex]Cl_2[/tex] in the reaction. Would all of the K be consumed, or would there still be excess K?
Moles of additional [tex]Cl_2[/tex] = mass of [tex]Cl_2[/tex] / molar mass of [tex]Cl_2[/tex]
Moles of additional [tex]Cl_2[/tex] = 0.50 g / 70.90 g/mol
Moles of additional [tex]Cl_2[/tex] = 0.0070 mol
The theoretical yield of KCl that can be formed from the additional [tex]Cl_2[/tex] is:
0.0070 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) x (74.55 g KCl / 1 mol KCl) = 1.04 g KCl
Therefore, the total amount of KCl that can be formed from all of the [tex]Cl_2[/tex] is:
5.95 g + 1.04 g = 6.99 g
The amount of K that would be needed to completely consume all of the [tex]Cl_2[/tex].
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Explain the term orbital.
Answer: a mathematical expression describing the probability of finding an electron at various locations; usually represented by the region of space around the nucleus where there is a high probability of finding an electron
Explanation: