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musickatia [10]
3 years ago
9

Determine the overall charge on each complex.a) tetrachlorocuprate(i)b) pentaamminechlorocobalt(iii)c) diaquadichloroethylenedia

minecobalt(iii)
Chemistry
1 answer:
balu736 [363]3 years ago
4 0
Complex a: <span>tetrachlorocuprate(i)
In present complex chlorine is negatively charged ligand (-1) and oxidation state of copper is +1. Therefore, total charge on complex = 4(-1) + 1 = -3

Complex b: </span><span>pentaamminechlorocobalt(iii)
In present complex, ammine is a neutral ligand (charge = 0), chlorine is negatively charged ligand (charge = -1) and oxidation state of Co is +3. Therefore, total charge on complex is 5(0) + (-1) + (+3) = +2

</span><span>Complex c: diaquadichloroethylenediaminecobalt(iii)
</span>In present complex, aqua is a neutral ligand (charge = 0), chlorine is a negatively charged ligand (charge = -1), ethylenediamine is a neutral ligand (charge = 0) and oxidation state of cobalt is +3. Therefore, total charge on complex is 2(0) + 2(-1) + 2(0) + 3 = +1. 
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<u>Answer:</u> The Henry's law constant for oxygen gas in water is 1.702\times 10^{-5}g/mL.atm

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To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{O_2}=K_H\times p_{O_2}

where,

K_H = Henry's constant = ?

C_{O_2} = solubility of oxygen gas = 0.0043g/120.3mL

p_{O_2 = partial pressure of oxygen gas = 2.1 atm

Putting values in above equation, we get:

0.0043g/120.3mL=K_H\times 2.1atm\\\\K_H=\frac{0.0043g}{120.3mL\times 2.1atm}=1.702\times 10^{-5}g/mL.atm

Hence, the Henry's law constant for oxygen gas in water is 1.702\times 10^{-5}g/mL.atm

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