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4 years ago

Select the correct value for the indicated bond angle in each of the following compounds:

Chemistry

2 answers:

serg [7]4 years ago

4 0

For O3, it will have an angle of <120. There are three bonding domains on the central O. 2 bonds and one lone pair. So you start with a bond angle of 120, but the lone pair takes up a little more space than the bonds. so the angle is a little less than 120.

SCl2 will be <109.5. There are 4 bonding domains on S (2 bonds and 2 lone pairs). So the angle starts at 109.5 but due to lone pairs, it is less (similar to a water molecule) 

SOCl2 will be <109.5 similar to SCl2 but 3 bonds and one lone pair on eh central atom.

azamat4 years ago

3 0

Answer:

SCl2 will be <109.5. There are 4 bonding domains on S (2 bonds and 2 lone pairs). So the angle starts at 109.5 but due to lone pairs, it is less (similar to a water molecule)

SOCl2 will be <109.5 similar to SCl2 but 3 bonds and one lone pair on eh central atom.

Explanation:

You might be interested in

How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group

OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

P= 1.00 atm
V= 150 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 $\frac{atm*L}{mol*K}$ *273 K

Solving:

$n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}$

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

$mass=\frac{6.69 moles*70.9 grams}{1 mole}$

mass= 474.321 grams ≅ 474 grams

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

4 0

3 years ago

What are the metals nonmetals and metalloids on periodic table?

deff fn [24]

Answer:

he line begins at boron (B) and extends down to polonium (Po). Elements to the left of the line are considered metals. Elements just to the right of the line exhibit properties of both metals and nonmetals and are termed metalloids or semimetals. Elements to the far right of the periodic table are nonmetals.

4 0

3 years ago

You will mostly find me in solid form at room temperature.

AlexFokin [52]

Nonmetal is the correct answer

5 0

3 years ago

A sample consisting of n mol of an ideal gas undergoes a reversible isobaric expansion from volume Vi to volume 3Vi. Find the ch

kolezko [41]

Answer:

The change in entropy of gas is $\Delta S= nC_{P}ln3$

Explanation:

n= Number of moles of gas

Change in entropy of gas = $ds= \int \frac{dQ}{T}$

$dQ= nC_{p}dT$

From the given,

$V_{i}=V$

$V_{f}=3V$

Let "T" be the initial temperature.

$\frac {V_{i}}{T_{i}}=\frac {V_{f}}{T_{f}}$

$\frac {V}{T}=\frac {3V}{T_{f}}$

${T_{f}} = 3T$

$\int ds = \int ^{T_{f}}_{T_{i}} \frac{nC_{P}dT}{T}$

$\Delta S = nC_{p}ln(\frac{T_{f}}{T_{i}})$

$\Delta S = nC_{p}ln3$

Therefore, The change in entropy of gas is $\Delta S= nC_{P}ln3$

3 0

3 years ago

A 360. mg sample of aspirin, C9H8O4, (molar mass 180. g), is dissolved in enough water to produce 200. mL of solution. What is t

Gemiola [76]

Answer:

Molarity = 0.01 M

Explanation:

Molarity is used to measure the concentration of a solution. It will be same for the whole solution or a small amount of solution if the solution is homogeneous.

So, Molarity of 200 mL of solution = Molarity of 50 mL of solution

$\mathbf{Molarity = \frac{number \ of \ moles \ of \ solute}{Volume \ of \ solution \ (in \ liters)}}$