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Feliz [49]
3 years ago
9

While in Europe, if you drive 109 km per day, how much money would you spend on gas in one week if gas costs 1.10 euros per lite

r and your car's gas mileage is 22.0 mi/gal ? Assume that 1euro=1.26dollars.
Chemistry
1 answer:
MrRissso [65]3 years ago
5 0

Answer:

The amount is x  =  113.3 \  dollars

Explanation:

From the question we are told that

     The  distance traveled per day is  l  =  109\ km =  \frac{109}{1.609}  =  67.74 \  mi

    The  cost of one liter is  c =  1.10 \ euros/liter = 1.10  *  1.26  = 1.36 \ dollars/liter

     The car's  gas mileage is  b =  22.0 \ mi/gal

Generally the amount of distance covered in one week is evaluated as

      z =  67.74 * 7

       z = 474.18 \  mi

The  amount of gas used in one week by the car is mathematically represented as  

       k  = \frac{ z}{ b}

=>    k  = \frac{ 474.18}{ 22}

=>      k  =  22 \ gal

converting to liters

          k =  22 *  3.78541=83.28 \ liters

Thus the amount spent on gas in one week is  

          x =  k *  c

=>      x  =  83.28 *  1.36

=>      x  =  113.3 \  dollars

 

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A solution of Na2CO3 is added dropwise to a solution that contains 1.15×10−2 M Fe2+ and 0.58×10−2 M Cd2+. What concentration of
castortr0y [4]

The question is incomplete, complete question is;

A solution of Na_2CO_3 is added dropwise to a solution that contains1.15\times 10^{-2} M of Fe^{2+} and 0.58\times 10^{-2} M and Cd^{2+}.

What concentration of CO_3^{2-} is need to initiate precipitation? Neglect any volume changes during the addition.

K_{sp} value FeCO_3: 2.10\times 10^{-11}

K_{sp} value CdCO_3: 1.80\times 10^{-14}

What concentration of CO_3^{2-} is need to initiate precipitation of the first ion.

Answer:

Cadmium carbonate will precipitate out first.

Concentration of CO_3^{2-} is need to initiate precipitation of the cadmium (II) ion is 3.103\times 10^{-12} M.

Explanation:

1) FeCO_3\rightleftharpoons Fe^{2+}+CO_3^{2-}

The expression of an solubility product of iron(II) carbonate :

K_{sp}=[Fe^{2+}][CO_3^{2-}]

2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]

[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}

[CO_3^{2-}]=1.826\times 10^{-9}M

2) CdCO_3\rightleftharpoons Cd^{2+}+CO_3^{2-}

The expression of an solubility product of cadmium(II) carbonate :

K_{sp}=[Cd^{2+}][CO_3^{2-}]

1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]

[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}

[CO_3^{2-}]=3.103\times 10^{-12} M

On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.

So, cadmium carbonate will precipitate out first.

And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the 3.103\times 10^{-12} M concentration.

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