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3 years ago

Assuming that all infrared vibrations are active, what is

Chemistry

1 answer:

777dan777 [17]3 years ago

7 0

Answer:

The number of lines possible for SO2 is 3

Explanation:

The following Procedure should be followed when calculating the number of vibrational modes:-

Identify if the given molecule is either linear or non-linear
Calculate the number of atoms present in your molecule
Place the value of n in the formula and solve.

SO2 is a non-linear molecule because it contains a lone pair which causes the molecule to bent in shape hence, The mathematical formula for calculating the number of non-linear molecule in a infrared region is (3n - 6) here n is the number of atoms in molecule.

hence for Sulphur Dioxide (SO2), n will be 3

Therefore, The number of lines possible for SO2 is (3*3) - 6 = 3

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0. Give an example of how an organism can use surface tension.

Lady_Fox [76]

Answer:

In case of plants surface tension help to support the transpiration pull.

Explanation:

Surface tension is a force that hold the molecules in the surface to minimize the surface area.

During evaporation of excess amount of water from the stomata of leaves of plants by transpiration a surface tension is generated.

The generated surface tension helps to maintain the water column within the xylem tissue by the absorption of water from the soil by the roots.

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3 years ago

The type of compound that is most likely to contain a covalent bond is ________.

max2010maxim [7]

The type of substance that is most likely to contain a covalent bond is ONE THAT IS COMPOSED OF ONLY NON METALS.
Covalent bond is a type of chemical bond in which electron pairs are shared among the participating elements in order to achieve the octet form. Covalent bond is usually found among non metals.

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3 years ago

A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

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3 years ago

How to know the elements amount of charge amount

MrRa [10]

For a single atom, the charge is the number of protons minus the number of electrons

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2 years ago

For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

nata0808 [166]

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

$\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)$ .

When $\rm H_2$ is added to $\rm H_2C\text{=}CH_2$ (ethene) under heat and with the presence of a catalyst, $\rm H_3C\text{-}CH3$ (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, $\Delta S < 0$ .

The equation for the change in Gibbs Free Energy for a particular reaction is:

$\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})$ .

For a particular reaction, the more negative $\Delta G$ is, the more spontaneous ("favorable") the reaction would be.

Since typically $\Delta S < 0$ for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

$T$ (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

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3 years ago