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stira [4]
2 years ago
11

Which of the following reactions would have the smallest value of K at 298 K? Which of the following reactions would have the sm

allest value of K at 298 K? A + B → 2 C; E°cell = -0.030 V A + 2 B → C; E°cell = +0.98 V A + B → C; E°cell = +1.22 V A + B → 3 C; E°cell = +0.15 V More information is needed to determine.
Chemistry
1 answer:
Firdavs [7]2 years ago
7 0

Answer:

The reaction with smallest value of K is :

A + B → 2 C; E°cell = -0.030 V

Explanation:

nFE^o_{cell}=RT\ln K

where :

n = number of electrons transferred

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = 25^oC=[273+25]=298K

K = equilibrium constant of the reaction

As we cans see, that standard electrode potential of the cell is directly linked to the equilibrium constant of the reaction.

  • Higher E^o_{cell} higher will be the value of K.
  • Lower E^o_{cell} lower will be the value of K.

So, the reaction with smallest value of electrode potential will have smallest value of equilibrium constant. And that reaction is:

A + B → 2 C; E^o_{cell} =-0.030 V

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A 1.44-g sample of an unknown pure gas occupies a volume of 0.335 L at a pressure of and a temperature of 100.0°C. The unknown g
Zepler [3.9K]

Answer:

Xenon

Explanation:

Step 1: Given data

  • Mass (m): 1.44 g
  • Volume (V): 0.335 L
  • Pressure (P): 1.00 atm (I looked it up)
  • Temperature (T): 100.0°C

Step 2: Convert the temperature to Kelvin

K = °C + 273.15 = 100.0°C + 273.15 = 373.2 K

Step 3: Calculate the number of moles (n)

We will use the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.00 atm × 0.335 L / (0.0821 atm.L/mol.K) × 373.2 K

n = 0.0109 mol

Step 4: Calculate the molar mass of the gas

M = 1.44 g / 0.0109 mol = 132 g/mol

Step 5: Identify the gas

The gas with a molar mass of about 132 g/mol is xenon.

8 0
2 years ago
Part 1: Name the type of chemical reaction that occurs when magnesium chloride solution (MgCl2) reacts with sodium carbonate sol
garri49 [273]

Answer:

Mg would blow off. AI would be affective to copper but not to MG

Explanation:

7 0
2 years ago
If you have ever peeled the label off a glass jar, you may have noticed that the glue does not easily wash off with water. Howev
shepuryov [24]

Answer:

It can be removed by acidic chemicals

Explanation:

8 0
2 years ago
A sample of an alloy of aluminum contains 0.0898 mol Al and 0.0381 mol Mg. What are the mass percentages of Al and Mg in the all
blondinia [14]

Answer:

Al 72.61%

Mg 27.39%

Explanation:

To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.

If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.

To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.

Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g

The mass of magnesium is 0.0381 * 24 = 0.9144g

We can now calculate the mass percentage.

The total mass is 0.9144 + 2.4246 = 3.339g

% mass of Al = 2.4246/3.339 * 100 = 72.61%

% mass of Mg = 0.9144/3.39 * 100 = 27.39%

7 0
2 years ago
Read 2 more answers
When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac
STatiana [176]

Answer:

\Delta _{comb}H=-2,093\frac{kJ}{mol}

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Q_{rxn}+Q_{cal}=0

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\  \Delta _{comb}H=-2,093\frac{kJ}{mol}

Best regards!

4 0
1 year ago
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