Answer is: <span>D)194 kJ/mol, exothermic
</span>ΔHreaction = ∑(product bond energies) - ∑(reactant bond energies).
∑(product bond energies) = ΔHreaction + ∑(reactant bond energies).
ΔHreaction must be negative (exothermic) if ∑(product bond energies) is positive.
∑(product bond energies) = -1352 kJ/mol + <span>1546 kJ/mol.
</span>∑(product bond energies) = 194 kJ/mol.
∑ is summation.
Answer:C.) It has one group in which the manipulated variable is tested, and another in which the responding variable is tested
Explanation: hope this helps
D because O3 means oxygen times 3 times it self so you add another oxygen partial it becomes O4
Answer : The heat energy absorbed will be, 
Solution :
The process involved in this problem are :
The expression used will be:
![\Delta H=m\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3Dm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
m = mass of ice = 1100 g
= specific heat of liquid water = 
= enthalpy change for fusion = 
Molar mass of water = 18 g/mole
Now put all the given values in the above expression, we get:
![\Delta H=1100g\times 333.89J/g+[1100g\times 4.18J/g^oC\times (32.0-0)^oC]](https://tex.z-dn.net/?f=%5CDelta%20H%3D1100g%5Ctimes%20333.89J%2Fg%2B%5B1100g%5Ctimes%204.18J%2Fg%5EoC%5Ctimes%20%2832.0-0%29%5EoC%5D)
Conversion used : (1 cal = 4.184 J)
Therefore, the heat energy absorbed will be,
Answer: gas and alkaline earth metal
Explanation:
