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3 years ago

Which of the following molecules would you expect to have a dipole moment of zero?

Chemistry

1 answer:

Mashcka [7]3 years ago

6 0

Answer:

CCl2=CCl2

Explanation:

Recall that the individual dipole moments in a symmetrical molecule cancel out each other thereby making the overall molecule to be non polar.

CCl2=CCl2 contain individual polar bonds but the molecule is highly symmetrical thus its dipole cancel out and the molecule has no overall dipole moment, hence CCl2=CCl2 has a dipole moment of zero .

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What term is used to describe the building blocks of life

KATRIN_1 [288]

Answer:

Explanation:

cell

7 0

2 years ago

A 4.7-L sealed bottle containing 0.33 g of liquid ethanol,C₂H₆O, is placed in a refrigerator and reaches equilibrium with its va

Mama L [17]

The mass of ethanol present in the vapor is 8.8×10⁻²g. when liquid and vapor ethanol at equilibrium.

The volume of the bottle = 4.7 L

Mass of ethanol = 0.33 g

Temperature (T1) = -11 oC = 273-11 = 262 K

P1 = 6.65 torr

Now we will calculate the mole by applying the ideal gas equation:-

PV = nRT

Or, n = PV/RT

Where P is the pressure

T is the temperature

R is the gas constant = 0.0821 L atm mol-1K-1

V is the volume

Substituting the values of P, V, T, and R the mole of ethanol is calculated as:-

= 0.001913 mol C2H6

Conversion of the mole to gm

Molar mass of ethanol (M) = 46.07 g/mol

Mass of C2H6O =0.001913 mol C2H6O 46.07 g/mol = 0.088 = 8.8×10⁻²g.

Hence, the mass of ethanol present in the vapor is found to be 8.8×10⁻²g.

Learn more about mole here:-brainly.com/question/15374113

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8 0

1 year ago

Round off each number to the indicated number significant figures (sf):

qaws [65]

Answer :

(a) The given number in 2 significant figures will be, $3.6\times 10^3$

(b) The given number in 4 significant figures will be, $3.583\times 10^1$

(c) The given number in 3 significant figures will be, $2.24\times 10^1$

Explanation :

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

(a) The given number 00003554 converted into 2 significant figures.

In 00003554, there are 4 significant figures. Now we have to convert it into 2 significant figures.

The given number in 2 significant figures will be, $3.6\times 10^3$

(b) The given number 35.8348 converted into 4 significant figures.

In 35.8348, there are 6 significant figures. Now we have to convert it into 4 significant figures.

The given number in 4 significant figures will be, $3.583\times 10^1$

(c) The given number 22.4555 converted into 3 significant figures.

In 22.4555, there are 6 significant figures. Now we have to convert it into 3 significant figures.

The given number in 3 significant figures will be, $2.24\times 10^1$

7 0

2 years ago

Select the following statements that are true. Select all that apply.

geniusboy [140]

Answer:

The statements which are true among these are: (a),(b) and (c) because,

(a) The simplest organic compounds which contains only carbon and hydrogen atoms are called hydrocarbons.

(b) The IUPAC naming of organic compounds have some rules for the naming of compounds, which consists of

Finding the longest chain present in the compound called parent chain.
A prefix for any substituent attach to the parent chain.

And lastly a suffix for the type of bond that molecule have.

(c) Isomers are the compound which same same molecular formula but different arrangement of molecules, due to this different arrangement they have different physical and chemical properties.

4 0

3 years ago

g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron

Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ → Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0

3 years ago