Answer:
The distance in kilometer that Ted will go if he runs at the same pace for 285 minutes is 60km
Step-by-step explanation:
It was given that Ted Ted runs 20 km in 95 minutes, so for us to know how many kilometers he runs at the same pace for 285 minutes by using proportion.
For Ted to run 20 km in 95 minutes.
where given distance= 20km
Given time =95minutes
Speed=20/95 .............eqn(1)
To calculate the distance in kilometers Ted will go if he runs at the same pace for 285 minutes, then
Speed= K/225 ...............eqn(2)
Let us represent distance with 'K'
Given time 225minutes
Equating eqn(1) and eqn(2)
Hence the proportion model that Ted could use in modelling this situation is this;
Cross multiply
K×95=20×225
The distance in kilometer that Ted will go if
he runs at the same pace for 285 minutes is 60km
The correct answer for the question that is being presented above is this one: "<span>16.728 g."</span>
Given that
ΔHsolid = -5.66 kJ/mol.
This means that 5.66 kJ of heat is released when 1 mole of NH3 solidifies
When 5.57 kJ of heat is released
amount of NH3 solidifies = 5.57/5.66 = 0.984 moles
<span>molar mass of NH3 = 17 g/mole </span>
<span>1 mole of NH3 = 17 g </span>
So, 0.984 moles of NH3 = 17 X 0.984 = 16.728 g
Data:
Q (Amount of heat) = 832 J
m (mass) = ?
c (Specific heat) = <span>0.90 J/(g × ° C)
T (final) = 97 ºC
To (initial) = 20 ºC
</span>ΔT = T - To → ΔT = 97 - 20 → ΔT = 77 ºC
Formula:
Q = m*c*ΔT
Solving:
Q = m*c*ΔT
832 = m*0.90*77
832 = 69.3m
69.3m = 832
Wave A has the shorter wavelength due too the fact that the shorter the wavelength the higher the frequency as they have an inverse relationship.
The classic Periodic Table<span> organizes the chemical </span>elements<span> according to the </span>number of<span> protons that each has in its atomic nucleus. Hope this helped :)</span>
