Answer:
The speed of the 60.0 kg skater should be 0.281 m/s
Explanation:
<u>Step 1: </u>Data given
Mass of skater 1 = 45.0 kg
speed of skater 1 = 0.375 m/s
Mass of skater 2 = 60.0 kg
<u>Step 2:</u> Calculate the speed of skater 2
To solve this problem, we will use 'Conservation of momenton'. This means the momentum before the push equals the momentum after.
momentum p = m*v
Momentum p(before) = momentum p(after)
m1*v1 = m2 * v2
⇒ with m1 = mass of skater 1 = 45.0 kg
⇒ with v1 = the velocity of skater 1 = 0.375 m/s
⇒ with m2 = the mass of skater 2 = 60.0 kg
⇒ with v2 = the velocity of skater 2 = TO BE DETERMINED
45.0 * 0.375 = 60.0 * v2
v2 = (45.0*0.375)/60
v2 = 0.281 m/s
The speed of the 60.0 kg skater should be 0.281 m/s
The answer is (3) 0.200 kJ
Hope I helped!
Given :
Mass of
is 571.6 g per liter .
Density of solution ,
.
To Find :
a. Mass percentage
b. Mole fraction
c. Molality
d. molarity of H2SO4 in this solution.
Solution :
Molar mass of
, m = 1329 g/mol .
a ) Mass of
contain in 1 liter is 1329 g .

b ) Moles of
=
.
Moles of
=
.
Mole fraction
.
c ) Molarity of
.
Hence , this is the required solution .
What don’t understand question
there should be less liquid at the end of the experiment since heating liquid is an endothermic process, and endothermic processes typically break the bonds between liquids. The applied heat would soon break the bonds that hold the molecules of the liquid together, and evaporation would occur once the temperature is hot enough. Thus, the evaporated water would decrease the volume of the liquid.