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)
Answer:
It would probably be, something that can take up moisture to test it.
Explanation:
(to see if it can evaporate)
Answer:
The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)
The number of moles = 5 g / 132.14 g/mol = 0.038 mol
The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23
the number of positive ions present in the ammonium sulphate solution:
2 positive ions for every 1 molecule of (NH₄)₂SO₄
so 2 x 2.29x10^23 = 4.58x10^23
the number of negative ions present in the ammonium sulphate solution
1 negative ion for every 1 molecule of (NH₄)₂SO₄
so 1 x 2.29x10^23 = 2.29x10^23
the total number of ions present in the ammonium sulphate solution
4.58x10^23 + 2.29x10^23 = 6.87x10^23
The correct answer is option C, 5.02 x 10²² carbon atoms
Atomic mass of C = 12 g/mol
According to Avogadro, 1 mole of C has 6.023 x 10²³C atoms
Now 1 mole of C is equal to 12 g
Therefore, 12 g of C = 6.023 x 10²³ C atoms
1 g of C = 
C atoms = 5.02 x 10²² C atoms
A chemical bond is <span> lasting attraction between atoms that enables the formation of </span>chemical <span>compounds. </span>
