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Amanda [17]
1 year ago
7

Determine the electron-group arrangement, molecular shape, and ideal bond angle(s) for each of the following:

Chemistry
1 answer:
Bad White [126]1 year ago
5 0

The electron group arrangement of NO²⁻is trigonal planar. The molecular shape is bent, and the bond angle is 120°.

<h3>What is the molecular shape of a compound?</h3>

The molecular geometry of the compound shows the position of nuclei and the electron of the compound. It shows how the joining of electrons and nuclei makes the shape of the compound.

Like here, the shape of nitrite is bent with lone pair which is shown by Lewis's structure The bond angle will be the distance between the nuclei of the neighbor atoms.

Thus, the electron geometry arrangement of nitrite is trigonal planer with a bent shape and the bond angle will be 120°.

To learn more about molecular geometry, refer to the link:

brainly.com/question/7558603

#SPJ4

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Intensive properties do not depend on the quantity of matter. Examples include density, state of matter, and temperature. Extensive properties do depend on sample size. Examples include volume, mass, and size.

Explanation:

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3 years ago
onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

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