Answer:Carbohydrates improve athletic performance by delaying fatigue and allowing an athlete to compete at higher levels for longer.
Explanation:
The answr to your question is 325.15
Answer: The specific heat capacity of chromium is 
Explanation:
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of chromium = 15.5 g
= mass of water = 55.5 g
= final temperature =
= temperature of chromium = 
= temperature of water = 
= specific heat of chromium= ?
= specific heat of water= 
Now put all the given values in equation (1), we get
![-15.5\times c_1\times (19.5-18.9)=[55.5\times 4.184\times (19.5-100)]](https://tex.z-dn.net/?f=-15.5%5Ctimes%20c_1%5Ctimes%20%2819.5-18.9%29%3D%5B55.5%5Ctimes%204.184%5Ctimes%20%2819.5-100%29%5D)
The specific heat capacity of chromium is
Colligative properties calculations are used for this type of problem. Calculations are as follows:
ΔT(boiling point) = 101.02 °C - 100.0 °C= 1.02 °C
<span>ΔT(boiling point) = (Kb)m
</span>m = 1.02 °C / 0.512 °C kg / mol
<span>m = 1.99 mol / kg
</span><span>ΔT(freezing point) = (Kf)m
</span>ΔT(freezing point) = 1.86 °C kg / mol (<span>1.99 mol / kg)
</span>ΔT(freezing point) = 3.70 <span>°C
</span>Tf - T = 3.70 <span>°C
T = -3.70 </span><span>°C</span>
