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masya89 [10]
2 years ago
14

Denise owns a car that runs on a mixture of gasoline and ethanol. She can buy fuels that have 85% ethanol or 25% ethanol. How mu

ch of each type of fuel should she buy if she wants to fill her 20-gallon tank with a mixture of fuel that contains 50% ethanol?
Chemistry
1 answer:
Hunter-Best [27]2 years ago
5 0
<span>85% ethanol | 25% ethanol | 50% ethanol
x | y | 20 gal

use x and y because you don;t know how much she needs.

0.85x | 0.25y | 20(0.5)

85% is 85/100 or 0.85, and you need that much of x, same goes for the 25% and 50% mixtures so now you can make up 2 equations
1) x + y = 20 2) 0.85x + 0.25y= 10 (you get 10 when you multiply 20 by 0.5) now you can solve for x or y using substitution.
first rewrite 1) in terms of x or y: x+ y= 20 ----> y= 20 - x now you can substitute 20- x for y in the second equation.. 0.85x + 0.25y= 10 0.85x + 0.25(20-x)= 10 distribute here..(0.25 * 20 and 0.25 * (-x) ) 0.85x + 5 - 0.25x = 10 combine like terms 0.6x +5 = 10 move the 5 over to the other side 0.6x= 10 -5 0.6x = 5 divide both sides by 0.6 x= 25/3 or 8.3 now you know the amount of x so you can substitue this back into the first equation to find y. 0.85x + 0.25y= 10 0.85(25/3) +0.25y= 10 85/12 + 0.25y= 10 0.25y = 10- 85/12 0.25y= 35/12 y= 35/3 or 11.6 you can check by putting these values into the euations: 1) x+ y= 20 25/3 + 35/3 =20 20= 20 good so far 2) 0.85x + 0.25y= 10 0.85(25/3) + 0.25(35/3)=10 10 = 10

so our values for x and y work
x= 25/3 and y= 35/3</span>
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Answer:

hello your question is incomplete attached below is the complete question

Write the balanced equation for the formation of the Grignard reagent from bromobenzene. Include all reagents and products BUT NOT SOLVENTS.

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What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Ag(s) and Ni2+(aq) to give Ni(s) and A
rusak2 [61]

Answer: 3\times 10^{35}

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The balanced chemical equation will be:

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

Here Ag undergoes oxidation by loss of electrons, thus act as anode. Nickel undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0_{[Ag^{+}/Mg]}=+0.80V

E^0_{[Ni^{2+}/Ni]}=-0.25V

E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Ag^{+}/Ag]}

E^0=-0.25-(+0.80V)=-1.05V

The standard emf of a cell is related to Gibbs free energy by following relation:

\Delta G=-nFE^0

\Delta G = gibbs free energy  

n= no of electrons gained or lost  =?

F= faraday's constant

E^0 = standard emf

\Delta G=-2\times 96500\times (-1.05)=202650J

The Gibbs free energy is related to equilibrium constant by following relation:

\Delta G=-2.303RTlog K

R = gas constant = 8.314 J/Kmol

T = temperature in kelvin =25^0C=25+273=298K

K = equilibrium constant

\Delta G=-2.303RTlog K

+202650=-2.303\times 8.314\times 298\times logK

K=3\times 10^{35}

Thus the value of the equilibrium constant at 25^0C is 3\times 10^{35}

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3 years ago
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2 years ago
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Calculate the wavelength of the photon that would be absorbed or emitted. Round your answer to 3 significant digits.
Flauer [41]

This is an incomplete question, the image for the given question is attached below.

Answer : The wavelength of photon would be absorbed, 3.06\times 10^{-7}m

Explanation :

From the given diagram of energy we conclude that,

Energy at ground state, A = 400 zJ

Energy of 2nd excited state, C = 1050 zJ

Now we have to calculate the energy of the photon.

E=E_c-E_A

E=(1050-400)zJ= 650zJ=650\times 10^{-21}J

Now we have to calculate the wavelength of the photon.

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

E = energy of photon = 650\times 10^{-21}J

\nu = frequency of photon

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength of photon  = ?

c = speed of light = 3\times 10^8m/s

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

650\times 10^{-21}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}

\lambda=3.06\times 10^{-7}m

Therefore, the wavelength of photon would be absorbed, 3.06\times 10^{-7}m

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3 years ago
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