Answer:
pOH of resulting solution is 0.086
Explanation:
KOH and CsOH are monoacidic strong base
Number of moles of 
in 375 mL of 0.88 M of KOH = 
= 0.33 moles
Number of moles of in 496 mL of 0.76 M of CsOH = 
= 0.38 moles
Total volume of mixture = (375 + 496) mL = 871 mL
Total number of moles of in mixture = (0.33 + 0.38) moles = 0.71 moles
So, concentration of in mixture, ![[OH^{-}]](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D)
= 
Hence, ![pOH=-log[OH^{-}]=-log(0.82)=0.086](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E%7B-%7D%5D%3D-log%280.82%29%3D0.086)
Compound= many parts together therefore bike is the answer
Answer:
24 atoms makes altogether a molecule of glucose
Answer:
Option D. 4.02 kJ
Explanation:
A simple calorimetry problem
Q = m . C . ΔT
ΔT = Final T° - Initial T°
C = Specific heat capacity
m = mass
Let's replace the data
Q = 125 g . 2.42 J/g∘C . (34.8°C -21.5 °C)
Q= 4023.25 J
We must convert the answer to kJ
4023.25 J . 1kJ /1000 =4.02kJ
