Answer:
[tex]V_2=8.4L[/tex]
Explanation:
Hello there!
In this case, according to the definition of the Boyle's law, which describes de pressure-volume behavior as an inversely proportional relationship, it is possible for us to write:
[tex]P_1V_1=P_2V_2[/tex]
Thus, since we are given the initial pressure and temperature, and the final pressure, we are able to calculate the final volume as shown below:
[tex]baV_2=\frac{P_1V_1}{P_2}\\\\V_2=\frac{8.0L*1.013bar}{ 0.968bar}\\\\V_2=8.4L[/tex]
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explain why hydrogen chloride does not conduct electricity, but a solution of hydrogen chloride and water conduct electricity
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 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.
Find out more information about methane here:
https://brainly.com/question/24690118
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.
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
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
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:
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 .