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vesna_86 [32]
3 years ago
9

Calculate the standard potential, e∘, for this reaction from its equilibrium constant at 298 k. x(s)+y4+(aq)↽−−⇀x4+(aq)+y(s)k=4.

98×10−5
Chemistry
1 answer:
laila [671]3 years ago
6 0
First, we need to get the number of moles:

from the reaction equation when Y4+ takes 4 electrons and became Y, X loses 4 electrons and became X4+  

∴ the number of moles n = 4

we are going to use this formula:

㏑K = n *F *E/RT

when K is the equilibrium constant = 4.98 x 10^-5

and F is Faraday's constant = 96500

and the constant R = 8.314

and T is the temperature in Kelvin = 298 K

and n is number of moles of electrons = 4 

so, by substitution:

㏑4.98 x 10^-5 = 4*96500*E / 8.314*298

∴E = -0.064 V
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A sample of mercury has a mass of 607.0 lb and a volume of 0.717 ft. What is its specific gravity? Number SG= What are the units
stepan [7]

Answer:

Specific gravity of mercury is 13.56 and it is an unit-less quantity.

Explanation:

Mass of the mercury = m = 607.0 lb = 275330.344 g

1 lb = 453.592 g

Volume of the mercury  = v = 0.717 ft^3=20,303.18 mL

1 ft^3 = 28316.847 mL

Density of the mercury = d=\frac{m}{v}=\frac{275330.344  g}{20,303.18 mL}

d = 13.56 g/mL

Specific gravity of substance = Density of substance ÷ Density of water

S.G=\frac{d}{1 g/mL}

Specific gravity of mercury :

S.G=\frac{13.56 g/mL}{1 g/mL}=13.56 (unit-less quantity)

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3 years ago
How does the motion of objects with similar mass compare to objects<br> of different mass?
emmasim [6.3K]

Explanation:

<em><u>in fact , we can use newtons second law of motion (see the SPT: Force topic) to calculate the acceleration in each of these cases</u></em>

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2 years ago
A 3.00 L flexible container holds a sample of hydrogen gas at 153 kPa. If the pressure increases to 203 kPa and the temperature
dybincka [34]

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

 

P1V1 =P2V2

V2 = P1 V1 / P2

V2 = 153 x 3.00 / 203

<span>V2 = 2.26 L</span>

3 0
2 years ago
. An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend in altitude until the pressure is 0.
Tju [1.3M]

Answer:

The answer to your question is Volume = 11.4 L

Explanation:

Data

Volume 1 = V1 = 6 L

Pressure 1 = P1 = 1 atm

Temperature 1 = T1 = 22°C

Volume 2 = V2 = ?

Pressure 2 = 0.45 atm

Temperature 2 = -21°C

Process

1.- Convert temperature (°C) to °K

T1 = 273 + 22 = 295°K

T2 = 273 + (-21) = 252°K

2.- Use the combined gas law to solve this problem

                   P1V1 / T1 = P2V2 / T2

-Solve for V2

                   V2 = P1V1T2 / T1P2

-Substitution

                   V2 = (6)(1)(252) / (295)(0.45)

- Simplification

                   V2 = 1512 / 132.75

- Result

                   V2 = 11.38 L

3 0
2 years ago
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