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3 years ago

Depicted below you will see the molecules COH2 and SO3 . Why is COH2 polar while SO3 isn’t?

Chemistry

1 answer:

tia_tia [17]3 years ago

7 0

Answer:

See explanation

Explanation:

When we look at SO3 we will notice that the compound is trigonal planar and it is symmetric. This means that it has equal charge distribution hence its dipoles cancel out resulting in a zero net dipole moment.

However, COH2 is also trigonal planar but is non-symmetric. Hence, its dipole moments do not cancel out, hence the molecule has a resultant dipole moment and is a polar molecule

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A ____usually requires calculations that are too complicated to do by<br>hand.

liraira [26]

Answer:

computer model

Explanation:

Computer models are cheaper to set up than alternative methods that could be used to predict what will happen in a system, ex. building a prototype. Other benefits include being able to: make alterations and quickly see the outcomes.

8 0

4 years ago

ASAP PLEASEE

Agata [3.3K]

Seven diatomic elements are H₂, Cl₂, N₂, F₂, Br₂, I₂ and O₂.

<h3>Which are diatomic molecules?</h3>

Diatomic molecules are those molecules in which two atoms of same elements are present, and they are combined to attain the stability.

The seven diatomic molecules which are exist in the chemistry are:

Hydrogen gas (H₂)
Chlorine gas (Cl₂)
Nitrogen gas (N₂)
Fluorine gas (F₂)
Bromine gas (Br₂)
Iodine gas (I₂)
Oxygen gas (O₂)

Hence H₂, Cl₂, N₂, F₂, Br₂, I₂ and O₂ are 7 diatomic molecules.

To know more about diatomic molecules, visit the below link:
brainly.com/question/14466404

#SPJ1

7 0

2 years ago

A first-order reaction (A → B) has a half-life of 25 minutes. If the initial concentration of A is 0.900 M, what is the concentr

Oksanka [162]

Answer:

0.6749 M is the concentration of B after 50 minutes.

Explanation:

A → B

Half life of the reaction = $t_{1/2}=25 minutes$

Rate constant of the reaction = k

For first order reaction, half life and half life are related by:

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{25 min}=0.02772 min^{-1}$

Initial concentration of A = $[A]_o=0.900 M$

Final concentration of A after 50 minutes = $[A]=?$

t = 50 minute

$[A]=[A]_o\times e^{-kt}$

$[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}$

[A] = 0.2251 M

The concentration of A after 50 minutes = 0.2251 M

The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M

0.6749 M is the concentration of B after 50 minutes.

7 0

3 years ago

5. A penny weighs about 2.5 g. How many moles of pennies would be required to equal the mass of the moon (7.3x10^24 kg)

schepotkina [342]

5. 1.16 x $10^{26}$ moles moles of pennies would be required to equal the mass of the moon.

6. 12.86 moles of ethanol are in a 750 ml bottle of vodka.

Explanation:

5 .Data given:

mass of penny = 2.5 grams

atomic mass of penny = 62.93 grams/mole

moles present in mass of the moon given as = 7.3 x $10^{24}$ kg

number of moles = $\frac{mass}{atomic mass of 1 mole}$

number of moles = $\frac{2.5}{62.93}$

0.039 moles of penny is present in 2.5 grams

0.039 moles of penny in 2.5 grams of it

so, x moles in 7.3 X $10^{27}$ grams

$\frac{0.039}{2.5} =\frac{x}{7.3 X 10^{27} }$

x = 1.16 x $10^{26}$ moles

so when the mass of the penny given is equal to the mass of moon, number of moles of penny present is 1.1 x $10^{26}$ .

Given:

vodka = 40% ethanol

volume of vodka bottle = 750 ml

moles of ethanol =?

density of ethanol =0.79 g/ml

atomic mass of ethanol = 46.07 grams/mole

so, from the density of ethanol given we can calculate how much ethanol is present in the solution.

density = $\frac{mass}{volume}$

density x volume = mass

0.79 x 750 = 592.5 grams

number of moles = $\frac{mass}{atomic mass of 1 mole}$

number of moles of ethanol = $\frac{592.5}{46.07}$

= 12.86 moles of ethanol

6 0

3 years ago

In an experiment, a 0.5297 g sample of diphenylacetylene (C14H10) is burned completely in a bomb calorimeter. The calorimeter is

V125BC [204]

Answer:

the Molar heat of Combustion of diphenylacetylene $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

Explanation:

Given that:

mass of diphenylacetylene $(C_{14}H_{10})$ = 0.5297 g

Molar Mass of diphenylacetylene $(C_{14}H_{10})$ = 178.21 g/mol

Then number of moles of diphenylacetylene $(C_{14}H_{10})$ = $\frac{mass}{molar \ mass}$

= $\frac{0.5297 \ g }{178.24 \ g/mol}$

= 0.002972 mol

By applying the law of calorimeter;

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = Heat absorbed by $H_2O$ + Heat absorbed by the calorimeter

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = msΔT + cΔT

= 1369 g × 4.184 J g⁻¹°C⁻¹ × (26.05 - 22.95)°C + 916.9 J/°C (26.05 - 22.95)°C

= 17756.48 J + 2842.39 J

= 20598.87 J

Heat liberated by 0.002972 mole of diphenylacetylene $(C_{14}H_{10})$ = 20598.87 J

Heat liberated by 1 mole of diphenylacetylene $(C_{14}H_{10})$ will be = $\frac{20598.87 \ J}{0.002972 \ mol}$

= 6930979.139 J/mol

= 6930.98 kJ/mol

Since heat is liberated ; Then, the Molar heat of Combustion of diphenylacetylene $(C_{14}H_{10})$ = $-6.931 *10^3 \ kJ/mol$

3 0

3 years ago

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