Directions:

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

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

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

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

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

3 years ago

A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throw

insens350 [35]

Answer:

Explanation:

This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:

$[m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a$ which, in words, is

The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:

$[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a$

Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).

0 = 90.0v - 2.0 and

2.0 = 90.0v so

v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.

12 = .022t and

t = 550 seconds, which is the same thing as 9.2 minutes

7 0

3 years ago

You are doing an experiment to determine your reaction time. Your friend holds a ruler. You place your fingers near the sides of

olga2289 [7]

I'm guessing that you mean like this:
-- The ruler is held with zero at the bottom, and the centimeter markings
    increase as you go up the ruler.
-- You place your fingers with the ruler and the zero mark between them.
-- The number where you catch the ruler is the distance it has fallen.

Then, all we have to find is the time it takes for the ruler to fall 11.3 cm .

Here's the formula for the distance an object falls from rest
in a certain time:

Distance = (1/2) (gravity) (time)²

On Earth, the acceleration of gravity is 9.8 m/s².
So we can write ...

      11.2 cm = (1/2) (9.8 m/s²) (time)²
or
       0.112 meter = (4.9 m/s²) (time)²

Divide each side
by 4.9 m/s² : (0.112 m) / (4.9 m/s²) = time²

(0.112 / 4.9) sec² = time²

Square root
each side:    time = √(0.112/4.9 sec²)

= √ 0.5488 sec²

=    0.74 second (rounded)

4 0

3 years ago

Read 2 more answers

In 10 seconds a car accelerates 4m/s^2 to 50

DanielleElmas [232]

Answer:

10m/s

Explanation:

using the formula a=v-u

t