Chemistry, 16.09.2019 08:20 juliannabartra

Car q and z are put though an endurance test to see if they can travel at 120 km/h for 5.0 hours. each car has a 45-l fuel tank. car z must stop to refuel after traveling for 4.2 hours. car q, however, travels for 5.4 hours before running out of gas. for each car, calculate the average kilometers traveled for each liter of gas (km/l)

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Answer from: melmar76

Explanation:

For car Z, according to the Newton equation,

v = u + at

where, v = final velocity = 0

u = initial velocity = 120 km/h

a = acceleration

t = time = 4.2 hours

Putting the given values into the above formula as follows.

v = u + at

0 = $120 km/h + a \times 4.2 hr$

a = 28.571 $km/h^{2}$

Also, it is known that $v^{2} = u^{2} + 2as$

s = $\frac{v^{2} - u^{2}}{2a}$

Putting the values into the above formula calculate for 's' as follows.

s = $\frac{v^{2} - u^{2}}{2a}$

= $\frac{(0)^{2} - (120)^{2}}{2 \times 28.571 km/hr^{2}}$

= 252.0 km

Since, 45 L of fuel is used by the car to travel 252 km. Then calculate car traveled km per liter as follows.

$\frac{252 km}{45 L}$

= 5.6 km/L

For car Q, v = u + at

0 = $120 + a \times 5.4 hr$

a = $\frac{120 km/hr}{5.4 hr}$

= 22.22 $km/hr^{2}$

Now, calculate value of 's' for car Q as follows.

s = $\frac{v^{2} - u^{2}}{2a}$

= $\frac{(0)^{2} - (120)^{2}}{2 \times 22.22 km/hr^{2}}$

= 324.0 km

Hence, average kilometers traveled for liter of gas is as follows.

$\frac{324 km}{45 L}$

7.2 km/L

Thus, we can conclude that the average kilometers traveled for each liter of gas (km/L) by car Z is 5.6 km/L and for car Q is 7.2 km/L.

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

Answer from: Quest

Its the second one nonmetals are moldable and easily shaped

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Car q and z are put though an endurance test to see if they can travel at 120 km/h for 5.0 hours. ea...

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