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shepuryov [24]
2 years ago
8

A solution is prepared by adding 100 mL of 1.0 M HC2H3O2 (aq) to 100 mL of 1.0 M NaC2H3O2 (aq). The solution is stirred and its

pH is measured to be 4.73. After 3 drops of 1.0 M HCl are added to the solution, the pH of the solution is measured and is still 4.73. Which of the following equations represents the chemical reaction that accounts for the fact that acid was added but there was no detectable change in pH?(A) H3O+(aq) + OH−(aq) -> 2 H2O(l)
(B) H3O+(aq) + Cl−(aq) -> HCl(g) + H2O(l)
(C) H3O+(aq) + C2H3O2−(aq) -> HC2H3O2(aq) + H2O(l)
(D) H3O+(aq) + HC2H3O2(aq) -> H2C2H3O+(aq) + H2O(l)
Chemistry
1 answer:
DIA [1.3K]2 years ago
5 0

Answer:

(C) H3O+(aq) + C2H3O2−(aq) -> HC2H3O2(aq) + H2O(l)

Explanation:

A buffer is a solution of a weak acid and its salt. It mitigates against changes in acidity or alkalinity of a system. A buffer maintains the pH at a constant value by switching the equilibrium concentration of the conjugate acid or conjugate base respectively.

Addition if an acid shifts the equilibrium position towards the conjugate acid side while addition of a base shifts the equilibrium position towards the conjugate base side.

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Equal numbers of moles of He(g), Ar(g), and Ne(g) are placed in a glass vessel at room temperature. If the vessel has a pinhole-
mrs_skeptik [129]

Answer:

VP as function of time => VP(Ar) > VP(Ne) > VP(He).

Explanation:

Effusion rate of the lighter particles will be higher than the heavier particles. That is, the lighter particles will leave the container faster than the heavier particles. Over time, the vapor pressure of the greater number of heavier particles will be higher than the vapor pressure of the lighter particles.

=> VP as function of time => VP(Ar) > VP(Ne) > VP(He).

Review Graham's Law => Effusion Rate ∝ 1/√formula mass.

4 0
2 years ago
If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
2 years ago
You can identify a metal by carefully determining its density. A 8.44 g piece of an unknown metal is 1.25 cm long, 2.50 cm wide,
Dafna1 [17]

Answer: Aluminum, 2.70 g/cm^3

Explanation:

Density is defined as the mass contained per unit volume.  It is characteristic of a substance.

Density=\frac{mass}{Volume}

Given : Mass of object = 8.44 grams

Volume of object=length\times breadth\times height=1.25cm\times 2.50cm\times 1.0cm=3.125cm^3

Putting in the values we get:

Density=\frac{8.44g}{3.125cm^3}=2.70g/cm^3

Thus density of the object will be 2.70 g/cm^3  which matches that of aluminium.

4 0
2 years ago
Read 2 more answers
A compound contains 30. percent sulfur and 70. percent fluorine by mass. the empirical formula of the compound is
iogann1982 [59]
Assume 100 g of compound. This turns percent to mass. Calculate moles: 

S ---> 30/32 = 0.9375 
F ---> 70 / 19 = 3.6842 

get whole number ratio: 

0.9375 / 0.9375 = 1 
3.6842 / 0.9375 = 3.93 = 4 

Answer : SF4
7 0
2 years ago
Read 2 more answers
What volume does 4.53 moles of hydrogen at 1.78 atm and 301 K occupy
AleksAgata [21]
V = nRT / P = (4.53 mol) x (0.08205746 L atm/K mol) x (301 K) / (1.78 atm) = 62.9 L
5 0
2 years ago
Read 2 more answers
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