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3 years ago

7

A scooter maintains a pace of 1.45 minutes per mile, with fuel economy of 85 miles per gallon. How many seconds can it travel on

145 mL of fuel under these conditions

1 answer:

vlada-n [284]3 years ago

6 0

Answer:

281s

Explanation:

Given parameters:

Speed of the scooter = 1.45min per mile

Fuel economy of engine = 85miles per gallon

Quantity of fuel = 145mL

Unknown:

Time of travel with the volume of fuel given = ?

Solution:

To solve this problem, we need to find the distance the fuel will last.

Rate of fuel consumption by the engine = 85miles per gallon

Convert 145mL to gallons;

3785.41mL = 1 gallon

145mL will therefore give $\frac{145}{3785.41}$ = 0.038gallons

So;

Distance covered = 85miles per gallon x 0.038gallons = 3.23miles

From;

Rate of travel = $\frac{time }{distance}$

Time = rate of travel x distance = 1.45 minutes per mile x 3.23miles

Time = 4.7min

1 min = 60s

4.7min = 4.7 x 60 = 281s

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solniwko [45]

Answer:

110

Explanation:

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3 years ago

A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 degrees celsius,

Airida [17]

Answer:

0.714 liter.

Explanation:

Given:

The balloon initially has a volume of 0.4 liters and a temperature of 20 degrees Celsius.

It is heated to a temperature of 250 degrees Celsius.

Question asked:

What will be the volume of the balloon after he heats it to a temperature of 250 degrees Celsius ?

Solution:

By using:

$PV=nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 0.4 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

0.4 liter = $0.4\times10^{-3}=4\times10^{-4} m^{3}$

And initial temperature of balloon, $T_{1}$ = 20°C = (273 + 20)K

= 293 K

Let the final volume of balloon is $V_{2}$

And a given, final temperature of balloon, $T_{2}$ is 250°C = (273 + 250)K

= 523 K

Now, $V_{1}$ = $KT_{1}$

$4\times10^{-4}=K\times293\ (equation\ 1 )$

$V_{2}$ = $KT_{2}$

$=K\times523\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4\times10^{-4}}{V_{2} } =\frac{K\times293}{K\times523}$

K cancelled by K.

By cross multiplication:

$293V_{2} =4\times10^{-4} \times523\\V_{2} =\frac{ 4\times10^{-4} \times523\\}{293} \\ = \frac{2092\times10^{-4}}{293} \\ =7.14\times10^{-4}m^{3}$

Now convert it into liter with the help of calculation done above.

$7.14\times10^{-4} \times1000\\7.14\times10^{-4} \times10^{3} \\0.714\ liter$

Therefore, the volume of the balloon be after he heats it to a temperature of 250 degrees Celsius is 0.714 liter.

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3 years ago

The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

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3 years ago

Las fórmulas actuales de los compuestos propuestos por Dalton

KonstantinChe [14]

Answer:

La teoría atómica de Dalton fue el primer intento completo para describir toda la materia en términos de los átomos y sus propiedades.

Dalton basó su teoría en la ley de la conservación de la masa y la ley de la composición constante.

La primera parte de su teoría establece que toda la materia está hecha de átomos, que son indivisibles.

La segunda parte de su teoría establece que todos los átomos de un elemento dado son idénticos en masa y en propiedades.

La tercera parte de su teoría establece que los compuestos son combinaciones de dos o más tipos diferentes de átomos.

La cuarta parte de su teoría establece que una reacción química es un reordenamiento de átomos.

Partes de su teoría tuvieron que ser modificadas con base en el descubrimiento de las partículas subatómicas y los isótopos.

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3 years ago

A 7.47 g sample of calcium carbonate [CaCO3 (s)] absorbs 85 J of heat, upon which the temperature of the sample increases from 2

Alex_Xolod [135]

Answer:

Specific heat of calcium carbonate(C) = 0.82 (Approx)

Explanation:

Given:

Energy absorbs (q) = 85 J

Change in temperature (Δt) = 34.9 - 21 = 13.9°C

Mass of calcium carbonate = 7.47 g

Find:

Specific heat of calcium carbonate(C)

Computation:

Specific heat of calcium carbonate(C) = q / m(Δt)

Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)

Specific heat of calcium carbonate(C) = 85 / 103.833

Specific heat of calcium carbonate(C) = 0.8186

Specific heat of calcium carbonate(C) = 0.82 (Approx)

7 0

3 years ago