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diamong [38]
2 years ago
6

The compound Rh(CO)(H)(PH3)2 forms cis and trans isomers. Use this information

Chemistry
1 answer:
Thepotemich [5.8K]2 years ago
5 0

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.

Learn more: brainly.com/question/9616145

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renewable energy sources such as solar and wind DONT emit carbon dioxide and other greenhouse gases that contribute to global warming

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A sample of a gas is placed inside a cylinder that is a fixed size. The cylinder is heated as additional gas is added to it. Wha
makkiz [27]
The pressure of gas will increase because gaseous state is the final state and even if the heat added is evaporating some more gas is still added. It also depends on the temperature of heat added, if the temperature doesn't change the it's most likely for the pressure to be stable...
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4 0
3 years ago
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. What is the mass of 23.56 moles of Glucose<br> ples of Glucose - C6H1206?
GaryK [48]

Answer:

4244.48 g to the nearest hundredth.

Explanation:

The molar mass of Glucose = 6*12.011 + 12*1.008+ 6*15.999

= 180.156.

So 23.56 moles = 180.156 * 23.56 =  4244.48 g

8 0
3 years ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
2 years ago
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