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Alex73 [517]
3 years ago
8

If your car gets 30.3 mi/gal, how many gallons of gasoline would you use if you drove 506.3 miles? 1 mile = 5280 ft (exactly) In

clude one decimal place in your answer.
Chemistry
2 answers:
mr_godi [17]3 years ago
7 0

Answer : The number of gallons of gasoline needed are 16.7 gallons

Explanation :

As we are given that the car traveled 30.3 mi/gal. Now we have to determine the number of gallons of gasoline needed in 506.03 miles.

As, 30.3 miles car traveled in = 1 gallon

and, 1 miles car traveled in = \frac{1\text{ mile}}{30.3\text{mile}}\times 1 \text{ gallon}

So, 506.3 miles car traveled in = \frac{506.3\text{ mile}}{30.3\text{mile}}\times 1 \text{ gallon}

                                                    = 16.7 gallons

Therefore, the number of gallons of gasoline needed are 16.7 gallons

Salsk061 [2.6K]3 years ago
4 0

Answer: 16.7 gallon

Explanation:

Given: The car can drive 30.3 miles when 1 gallon of gasoline is used.

Distance covered = 506.3 miles

Thus for 30.3 miles, the amount of gasoline used= 1 gallon

For 506.3 miles, the amount of gasoline used=\frac{1}{30.3}\times 506.3=16.7gallon

Thus the amount of gasoline used is 16.7 gallons.


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Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c
uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]

4x^{2} = 81.9x^{2} -57.8x +9.66

77.9x^{2} -57.8x+9.66 = 0

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

[ICl]_{eq} = 2 ( 0.254) = 0.508 M

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0
3 years ago
How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? (answ
Jet001 [13]

Answer:

7.12 mm

Explanation:

From coulomb's law,

F = kqq'/r².................... Equation 1

Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.

Make r the subject of the equation,

r = √(kqq'/F).......................... Equation 2

Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute into equation 2

r = √[ (75×10⁻⁹ )²9.0×10⁹/1]

r = 75×10⁻⁹.√(9.0×10⁹)

r = (75×10⁻⁹)(9.49×10⁴)

r = 711.75×10⁻⁵

r = 7.12×10⁻³ m

r = 7.12 mm

Hence the distance between the point charge = 7.12 mm

3 0
3 years ago
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