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,
<span>What
is the ph of an acetic acid solution if 10 drops are titrated with 70
drops of a 0.65 m koh solution? (ka for acetic acid = 1.8 x 10-5)?
[KOH] = 0.65 M
[OH] = 0.65 M
</span>Dilute your mom
<span>[OH]Dil= 0.65 M * 70/80 = 0.56875 M
pH = 5.4
</span>
If we have 6.68% NaClO, it is the same as saying--> 6.68 grams NaClO= 100 mL of solution. we can use this as a conversion.
800. mL (6.68 mL/ 100 mL)= 53.4 mL
solution = solute + solvent
solute= NaClO
solvent= H2O
solvent= 800-53.4= 747 mL of H2O
so, we you need 53.4 mL of NaClO and 747 mL of water or 53.4 grams of NaClO and 747 mL of water
i believe its true bc ik for sure air is a homogenous mixture
