If the cyclist rode at an average speed of 10mph for 15 miles...
we can solve by dividing the distance by the speed to get time using the equation...
Δspeed = Δdistance / Δtime
Δtime = Δdistance / Δspeed
Δtime = 15 miles / 10 mph = 1.5 hours
Answer: measure the mass (48.425g) of KCl
Explanation:
To prepare the solution 0.65M KCl we must measure the mass of KCl that would be dissolved in 1L of the solution. This can be achieved by:
Molar Mass of KCl = 39 + 35.5 = 74.5g/mol
Number of mole (n) = 0.65
Mass conc of KCl = n x molar Mass
Mass conc of KCl = 0.65 x 74.5 = 48.425g
Therefore, to make 0.65M KCl, we must measure 48.425g
Answer:
<h2>500 J</h2>
Explanation:
The kinetic energy of an object can be found by using the formula

where
m is the mass
v is the velocity
From the question
m = 1.6 kg
v = 25 m/s
We have

We have the final answer as
<h3>500 J </h3>
Hope this helps you
Answer:
1.2 × 10⁴ cal
Explanation:
Given data
- Initial temperature: 80 °C
We can calculate the heat released by the water (
) when it cools using the following expression.

where
c is the specific heat capacity of water (1 cal/g.°C)

According to the law of conservation of energy, the sum of the heat released by the water (
) and the heat absorbed by the reaction (
) is zero.
