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3 years ago

Which one of the following molecules would be most polar?

1 answer:

4 0

This question is quite a controversial one, because three of the options represented above are very similar, but due to the fact that there is a HF, which is even more polar that water, it is getting clear. So, the right answer is definitely the first option, because <span>HF belongs to acids that are very polar compounds.</span>

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According to the arrhenius concept, if HNO3, were dissolved in water, it would act as?: a. Proton acceptor b. A base c. A source

Alexxandr [17]

Arrhenius concept just states that if a solution dissociates and forms H+ ions then its and acid, but if it dissociates and forms OH- than it's a base.

So the answer to your question would be D. An Acid

5 0

4 years ago

Read 2 more answers

What are aldehydes and ketones

kari74 [83]

Answer:

Aldehydes derive their name from the dehydration of alcohols. Aldehydes contain the carbonyl group bonded to at least one hydrogen atom. Ketones contain the carbonyl group bonded to two carbon atoms. Aldehydes and ketones are organic compounds which incorporate a carbonyl functional group, C=O

Explanation:

hope this helps

6 0

3 years ago

Classifying a carbon atom by the number of carbons to which it is bonded can also be done in more complex molecules that contain

defon

A sp³ hybridized carbon atom is the one which forms 4 single bonds with other atoms.

In the structure of Bilobalide, we can see 12 sp³ hybridized carbon atoms which are attached to other carbon, hydrogen and oxygen atoms.

In order to classify carbon atoms as 1°, 2°, 3°, or 4° we need to take a look at the atoms to which these carbon atoms are attached.

A 1° carbon atom is the one which is attached to only 1 carbon atom.

In the diagram, we can see 3 methyl groups ( -CH3) being attached to only 1 carbon atom. There is one more primary carbon atom.

There are 4 primary carbon atoms in Bilobalide.

A 2° carbon atom is the one which is attached to 2 other carbon atoms.

In Bilobalide we can see 4 secondary carbon atoms.

A 3° carbon atom is the one which is attached to 3 other carbon atoms.

In Bilobalide we can see only 1 tertiary carbon atom.

A 4° carbon atom is the one which is attached to 4 other carbon atoms.

In Bilobalide we can see 3 quaternary carbon atoms.

Please see the attached image for classification.

7 0

3 years ago

A chemist used 6.5 moles of water in this reaction. How many grams of water were used?

denpristay [2]

Answer:

$\boxed {\boxed {\sf 120 \ or \ 117 \ grams \ H_2O \ depending \ on \ significant \ figures }}$

Explanation:

We want to convert from moles of water to grams of water.

First, find the molar mass of water (H₂O) Look on the Periodic Table for the masses of hydrogen and oxygen.

Hydrogen (H): 1.008 g/mol
Oxygen (O): 15.999 g/mol

Next, add up the number of each element in water. The subscript of 2 comes after the H, so there are 2 moles of hydrogen.

2 Hydrogen: (1.008 g/mol*2) = 2.016 g/mol

Finally, add the molar mass of 2 hydrogen and 1 oxygen.

2.016 g/mol (2 Hydrogen) + 15.999 g/mol (1 oxygen)= 18.015 g/mol

Next, find the grams in 6.5 moles.

Use the molar mass we just found as a ratio.

$molar \ mass \ ratio: \frac{18.015 \ g \ H_2O}{1 \ mol \ H_2O}$

We want to find the grams in 6.5 moles. We can multiply the ratio above by 6.5

$6.5 \ mol \ H_2O * \frac{18.015 \ g \ H_2O}{1 \ mol \ H_2O}$

Multiply. Note that the moles of H₂O will cancel each other out.

$6.5 * \frac{18.015 \ g \ H_2O}{1}$

$6.5 * {18.015 \ g \ H_2O}$

$117.0975 \ g \ H_2O$

If we want to round to the technically correct significant figures, it would be 2 sig figs. The original measurement, 6.5, has 2 (6 and 5).

$\approx 120 \ g \ H_2O$

6 0

3 years ago

Given the following equations, determine the standard enthalpy of formation (ΔH∘f) for one mole of ICl3(g).I2(g)+3Cl2(g)⟶2ICl3(g

Natasha_Volkova [10]

The standard enthalpy of formation (ΔH°f) for one mole of ICl₃(g) is -88 kJ, as determined by Hess' law.

We want to determine the standard enthalpy of formation (ΔH°f) for one mole of ICl₃(g). The equation for which we are looking the enthalpy of reaction is:

0.5 I₂(s) + 1.5 Cl₂(g) ⟶ ICl₃(g)

We will use Hess' law, which states that the total enthalpy change during the complete course of a chemical reaction is independent of the number of steps taken. Let's consider the following thermochemical equations.

I₂(g) + 3 Cl₂(g) ⟶ 2 ICl₃(g) ΔH°298 = −214 kJ

I₂(s) ⟶ I₂(g) ΔH°298 = 38 kJ

We will add the reactions and their enthalpies. The resulting reaction is:

I₂(s) + 3 Cl₂(g) ⟶ 2 ICl₃(g) ΔH°298 = −-176 kJ

Finally, since we want to calculate the standard enthalpy per mole of ICl₃, we will divide the previous equation by 2.

0.5 I₂(s) + 1.5 Cl₂(g) ⟶ ICl₃(g) ΔH°298 = −88 kJ

The standard enthalpy of formation (ΔH°f) for one mole of ICl₃(g) is -88 kJ, as determined by Hess' law.

Learn more: brainly.com/question/22953190

6 0

2 years ago