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mario62 [17]
3 years ago
12

Consider the cell pt | h2 (1 atm); h+ (? m) || hg2cl2(s); cl− (1 m) | hg 2 h+ + 2 e − → h2 e 0 = 0.00 v hg2cl2 + 2 e − → 2 hg +

2 cl− e 0 = 0.268 v if the measured cell potential for the cell is 0.35 volts, what is the ph of the solution? 1. 1.39 2. less than 1.00 3. 4.74 4. 2.77 5. 5.45 020
Chemistry
2 answers:
tatyana61 [14]3 years ago
5 0

Answer:

pH ˜ 1.39

Explanation:

solution

given the cell equation as follows:

Pt | H2 (1 atm); H+ (? M) || Hg2Cl2(s); Cl- (1 M) | Hg

half of the reactions are

2H+ + 2 e- --> H2 (E0 = 0.00 V )

Hg2Cl2 + 2 e- --> 2Hg + 2Cl- (E0 = 0.268 V)

E0 cell = E0 cathode -E0 anode

= 0.268 V -0.00V ( the more of positive is cathode while the smaller is anode )

= 0.268V

knowing that  Ecell = E0 cell - 0.0592 V/n) log [oxd/red]

0.35 V = 0.268V -(0.0592V/2) log [H+ ]2 /[Cl- ]

0.35V-0.268V =- 0.0296V*2 log [H+ ]/[1.0]

0.082V = -0.0592V log [H+ ]

so 0.082V /0.0592V = - log [H+ ]

1.385 = - log [H+ ]

1.385 = pH (since pH = - log [H+ ] )

In conclusion, pH ˜ 1.39

Dmitry [639]3 years ago
3 0

Answer:

1.39

Explanation:

[Hg2Cl2]= 1M

[H^+] = ????

E°cell= 0.35V

E= 0.268 V

Therefore E for the reaction must -0.082 V

n= 2 moles of electrons

From Nernst Equation:

E= E°cell- 0.0592/n log [Red]/[Ox]

0.0268= 0.35- 0.0592/2 log 1/[Ox]^2

-0.082= -0.0296 log 1/[Ox]^2

log 1/[Ox]^2= 0.082/0.0296

log 1/[Ox]^2= 2.77

1/[Ox]^2=Antilog (2.77)

[Ox]^2=1.698×10^-3

[Ox] = 0.0412 M

But pH= -log [H^+]= -log(0.0412)= 1.385

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yawa3891 [41]

Answer:

3.10×10¯⁵ ft³.

Explanation:

The following data were obtained from the question:

Density (D) of lead = 11.4 g/cm³

Mass (m) of lead = 10 g

Volume (V) of lead =.?

Density (D) = mass (m) / Volume (V)

D = m/V

11.4 = 10 / V

Cross multiply

11.4 × V = 10

Divide both side by 11.4

V = 10 / 11.4

V = 0.877 cm³

Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:

1 cm³ = 3.531×10¯⁵ ft³

Therefore,

0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³

0.877 cm³ = 3.10×10¯⁵ ft³

Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.

Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.

7 0
3 years ago
4. (01.05 MC)
DochEvi [55]

Answer:

The correct option is;

The gas particles move faster, have the same molecular composition, and have weaker attractions between them than the liquid particles

Explanation:

The properties of the gas molecules in comparison to liquids are

1) The gas molecules are widely spread out

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3) A gas fills the container in which it is placed

For liquids

1) There are strong intermolecular forces holding the molecules together in a liquid

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5 0
3 years ago
Excess magnesium reacts with 165.0 grams of hydrochloric acid in a single displacement reaction.
JulsSmile [24]

Answer:

The volume of hydrogen gas produced will be approximately 50.7 liters under STP.

Explanation:

Relative atomic mass data from a modern periodic table:

  • H: 1.008;
  • Cl: 35.45.

Magnesium is a reactive metal. It reacts with hydrochloric acid to produce

  • Hydrogen gas \rm H_2, and
  • Magnesium chloride, which is a salt.

The chemical equation will be something like

\rm ?\;Mg\;(s) + ?\;HCl \;(aq)\to ?\;H_2 \;(g)+ [\text{Formula of the Salt}],

where the coefficients and the formula of the salt are to be found.

To determine the number of moles of \rm H_2 that will be produced, first find the formula of the salt, magnesium chloride.

Magnesium is a group 2 metal. The oxidation state of magnesium in compounds tends to be +2.

On the other hand, the charge on each chloride ion is -1. Each magnesium ion needs to pair up with two chloride ions for the charge to balance in the salt, magnesium chloride. The formula for the salt will be \rm MgCl_2.

\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ ?\;MgCl_2\;(aq).

Balance the equation. \rm MgCl_2 contains the largest number of atoms among all species in this reaction. Start by setting its coefficient to 1.

\rm ?\;Mg\;(s) + ?\;HCl\;(aq) \to ?\;H_2 \;(g)+ {\bf 1\;MgCl_2}\;(aq).

The number of \rm Mg and \rm Cl atoms shall be the same on both sides. Therefore

\rm {\bf 1\;Mg}\;(s) + {\bf 2\;HCl}\;(aq) \to ?\;H_2 \;(g)+ {1\;\underset{\wedge}{Mg}\underset{\wedge}{Cl_2}}\;(aq).

The number of \rm H atoms shall also conserve. Hence the equation:

\rm {1\;Mg}\;(s) + {2\;\underset{\wedge}{H}Cl}\;(aq) \to {\bf 1\;H_2 \;(g)}+ {1\;MgCl_2}\;(aq).

How many moles of HCl are available?

M(\rm HCl) = 1.008 + 35.45 = 36.458\;g\cdot mol^{-1}.

\displaystyle n({\rm HCl}) = \frac{m(\text{HCl})}{M(\text{HCl})} = \rm \frac{165.0\;g}{36.458\;g\cdot mol^{-1}} = 4.52576\;mol.

How many moles of Hydrogen gas will be produced?

Refer to the balanced chemical equation, the coefficient in front of \rm HCl is 2 while the coefficient in front of \rm H_2 is 1. In other words, it will take two moles of \rm HCl to produce one mole of \rm H_2. \rm 4.52576\;mol of \rm HCl will produce only one half as much \rm H_2.

Alternatively, consider the ratio between the coefficient in front of \rm H_2 and \rm HCl is:

\displaystyle \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}.

\displaystyle n(\text{H}_2) = n(\text{HCl})\cdot \frac{n(\text{H}_2)}{n(\text{HCl})} = \frac{1}{2}\;n(\text{HCl}) = \rm \frac{1}{2}\times 4.52576\;mol = 2.26288\;mol.

What will be the volume of that many hydrogen gas?

One mole of an ideal gas occupies a volume of 22.4 liters under STP (where the pressure is 1 atm.) On certain textbook where STP is defined as \rm 1.00\times 10^{5}\;Pa, that volume will be 22.7 liters.

V(\text{H}_2) = \rm 2.26288\;mol\times 22.4\;L\cdot mol^{-1} = 50.69\; L, or

V(\text{H}_2) = \rm 2.26288\;mol\times 22.7\;L\cdot mol^{-1} = 51.37\; L.

The value "165.0 grams" from the question comes with four significant figures. Keep more significant figures than that in calculations. Round the final result to four significant figures.

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