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MrMuchimi
2 years ago
5

A company that is concerned about the environment wants to encourage the use of their fuel cell cars, so they are putting

Chemistry
2 answers:
aleksley [76]2 years ago
8 0
Fossil cell cars reduce use of non renewable fossil fuels.
Juli2301 [7.4K]2 years ago
6 0

Answer:

B. Fuel cell cars reduce use of non-renewable fossil fuels

Explanation:

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How many molecules are there in 10.5 grams of iron (111) sulfate trihydrate? With work shown
MrRa [10]
Fe2(SO4)3.3H2O = 56*2 + 3*(32+16*4)+3*(1*2+16) = 454 g/mol
Mol =m/M = 10.5/454 = 0.023 mol
Can u mark it brainliest ?
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3 years ago
What source of thermal energy does an internal combustion engine use?
Dafna1 [17]

Answer:

D

Explanation:

The source of the thermal energy in a combustion engine is the spark from the spark plug that ignites the fuel and air mixture converting chemical energy into thermal energy.

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2 years ago
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What is creation in terms of chemical reactions?
shtirl [24]

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Products

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3 years ago
What is the molarity of H3PO4?
masya89 [10]
The answer is 14.8 hoped this helped
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2 years ago
Silver occurs in trace amounts in some ores of lead, and lead can displace silver from solution: Pb(s) + 2Ag+ (aq) LaTeX: \longr
VikaD [51]

Answer : The value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

Explanation :

The balanced cell reaction will be,

Pb(s)+2Ag^+(aq)\rightarrow Pb^{2+}(aq)+2Ag(g)

The half-cell reactions are:

Oxidation reaction (anode) : Pb(s)\rightarrow Pb^{2+}(aq)+2e^-

Reduction reaction (cathode) : 2Ag^+(aq)+2e^-\rightarrow 2Ag(g)

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

\Delta G^o = standard Gibbs free energy

F = Faraday constant = 96500 C

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = 0.93 V

Now put all the given values in the above formula, we get:

\Delta G^o=-2\times 96500\times 0.93

\Delta G^o=-179490J/mol=-179.49kJ/mol\approx -180kJ/mol

Now we have to calculate the value of 'K'.

\Delta G^o=-RT\ln K

where,

\Delta G_^o =  standard Gibbs free energy  = -180 kJ/mol

R = gas constant = 8.314\times 10^{-3}kJ/mole.K

T = temperature = 298 K

K = equilibrium constant = ?

Now put all the given values in the above formula 1, we get:

-180kJ/mol=-(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln K

K=3.6\times 10^{31}

Therefore, the value of \Delta G^o and K is, -180 kJ/mol and 3.6\times 10^{31}

5 0
3 years ago
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