The compound Rh(CO)(H)(PH3)2 forms cis and trans isomers. Use this information
1 answer:
The geometry of a complex is determined by the number of ligands that surround the central atom/ion. There are four ligands in the complex. The coordination number of the complex is 4. The oxidation number of Rh is zero. The preferred geometrical shape of this complex is square planar.
Complexes are formed when Lewis bases called ligands become attached to a central metal atom or ion. The coordination number of the complex refers to the number of ligands that surround the central metal atom/ion.
In this case, four ligands surround the central Rhodium atom in a zero oxidation state.
Since we were told that the compound exhibits cis/trans isomerism then it can not have a tetrahedral geometry. It must have a square planar geometry as tetrahedral complexes do not exhibit cis/trans isomerism.
The image of the cis- and trans- geometrical isomers of the compound is attached to this answer.
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The molarity of the NaOH solution is 0.03 M
We'll begin by calculating the mole of the KHP
- Mass = 0.212 g
- Molar mass = 204.22 g/mol
- Mole of KHP =?
Mole = mass /molar mass
Mole of KHP = 0.212 / 204.22
Mole of KHP = 0.001 mole
Next, we shall determine the molarity of the KHP solution
- Mole of KHP = 0.001 mole
- Volume = 50 mL = 50/1000 = 0.05 L
- Molarity of KHP =?
Molarity = mole / Volume
Molarity of KHP = 0.001 / 0.05
Molarity of KHP = 0.02 M
Finally , we shall determine the molarity of the NaOH solution
KHP + NaOH —> NaPK + H₂O
From the balanced equation above,
- The mole ratio of the acid, KHP (nA) = 1
- The mole ratio of base, NaOH (nB) = 1
From the question given above, the following data were obtained:
- Volume of acid, KHP (Va) = 50 mL
- Molarity of acid, KHP (Ma) = 0.02 M.
- Volume of base, NaOH (Vb) = 35 mL
- Molarity of base, NaOH (Mb) =?
MaVa / MbVb = nA / nB
(0.02 × 50) / (Mb × 35) = 1
1 / (Mb × 35) = 1
Cross multiply
Mb × 35 = 1
Divide both side by 35
Mb = 1 / 35
Mb = 0.03 M
Thus, the molarity of the NaOH solution is 0.03 M
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The answer for the following question is mentioned below.
<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>
Explanation:
Given:
Pressure of gas (P) = 1.2 atm
Volume of a gas (V) = 50.0 liters
Temperature (T) =650 K
To calculate:
no of moles present in the gas (n)
We know;
According to the ideal gas equation;
We know;
<u>P × V = n × R × T </u>
where,
P represents pressure of the gas
V represents volume of the gas
n represents no of the moles of a gas
R represents the universal gas constant
where the value of R is 0.0821 L atm mole^{-1} K^-1
T represents the temperature of the gas
As we have to calculate the no of moles of the gas;
n =
n = \frac{1.2*50.0}{0.0821*650}
n = \frac{60}{53.365}
n = 1.12 moles
<u><em>Therefore no of moles present in the gas are 1.12 moles</em></u>