Which is not a compound? water silicon dioxide oxygen gas carbon dioxide gas

If electrons were used in the two slit experiment instead of light, what change would need to be made to the slit spacing in ord

irina [24]

The change that would need to be made to the slit spacing in order to see a diffraction pattern is bending, because in understanding why light behaves like a wave, it is the interference and diffraction were the phenomena distinguish waves from particles but waves are the only one can interfere and diffract while particles do not. The light bends around obstacles or cylinder like waves do, then it is bending which cause and resulted in the single slit diffraction pattern.

5 0

3 years ago

Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit

luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

$K~=~10^{-pK}$

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

$K~=~10^{4.9}$

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

$K`=~10^{1.12]$

Reaction can take place

2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

$K~=~10^{ -3.56}$

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

$K`=~10 ^{-12.93}$

Reaction does not happen

3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

$K~=~10^{-3.26}$

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

$K~=~ 10^{-12.63}$

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

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the concentrations at equilibrium.

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the ph of a solution

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Keywords : acid base reaction, the equilibrium constant

5 0

3 years ago

Read 2 more answers

Saturated fatty acids and unsaturated fatty acids differ in.

ratelena [41]

Answer:

the number off double bonds in a fatty acid chain

4 0

1 year ago

Crude petroleum is a mixture of many different components. How might one separate out the components to begin the process of ref

dimaraw [331]

Fractional distillation is used for the refining of crude petroleum.

<h3>How are the components of crude petroleum separated out?</h3>

Fractional distillation is the procedure used to separate crude oil's numerous constituents.

A mixture is divided into several components, known as fractions, using fractional distillation.
A combination of hydrocarbons makes up crude oil. The crude oil evaporates, and in the fractionating column, its vapors condense at various temperatures.
The hydrocarbon molecules in each percent have a comparable number of carbon atoms and a comparable range of boiling points.
The mixture is placed above a tall fractionating column that has multiple condensers coming off at various heights.
The bottom of the column is warm, while the top is cool. High boiling point compounds condense at the bottom, whereas low boiling point substances condense as they ascend.

Learn more about fractional distillation here:

brainly.com/question/15187318

#SPJ4

8 0

1 year ago

Match the prefixes. 1. 1 hex- 2. 2 eth- 3. 3 prop- 4. 4 hept- 5. 5 non- 6. 6 dec- 7. 7 pent- 8. 8 but- 9. 9 meth- 10. 10 oct-

erma4kov [3.2K]

Hi!

All these prefixes are commonly employed in the nomenclature in organic chemistry. These are associated with organic compounds, such as alkanes, alkenes, alcohols and carboxylic acids -and these prefixes are known as word root.

<h3>The answers would be:</h3><h3>a) 1 - meth</h3>

Meth is associated with the number one, in that it signifies the presence of one carbon in the compound in concern. For instance, a one carbon alkane would be known as methane -with <em>ane </em>being indicative of the compound being an alkane, and meth denoting the presence of one carbon

Eth is the associated prefix with the number 2 as it denotes the presence of 2 carbon atoms in the compound. For instance, the compound ethene gets it name for having 2 carbon atoms (denoted by eth). The eth acts as a prefix to the characteristic <em>ene</em> of alkenes.

Prop is the prefix that signifies the presence of 3 carbons in the compound in concern. For instance, propanol is a three carbon alcohol, which we can tell by the name. Prop indicating three carbons in the compound, and <em>anol</em> being indicative of the compound being an alcohol.

But is associated with the number one, in that it denotes the presence of 4 carbons in the compound. For instance, a 4 carbon carboxylic acid would be known as butanoic acid -with anoic<em> </em>being indicative of the compound being a carboxylic acid, and but signifying the presence of four carbons.

Pent is the associated prefix with the number five as it denotes the presence of five carbon atoms in the compound. For instance, the compound pentene gets it name for having five carbon atoms (denoted by pent). The pent acts as a prefix to the characteristic <em>ene</em> of alkenes.

Hex is the prefix that signifies the presence of six carbons in the compound in question. For instance, a six carbon alkane would be known as hexane -with <em>ane </em>being indicative of the compound being an alkane, and hex denoting the presence of one carbon

Hept is the associated prefix with the number seven as it denotes the presence of seven carbon atoms in the compound. For instance, the compound heptene gets it name for having seven carbon atoms (denoted by hept). The hept acts as a prefix to the characteristic <em>ene</em> of alkenes.

Oct is the prefix that is associated with the number 8. This is because it denotes the presence of eight carbon atoms in the compound. For instance, the compound octane gets it name for having eight carbon atoms (denoted by oct). The oct acts as a prefix to the characteristic <em>ane</em> of alkanes.

Non is the prefix that is associated with the number nine, and thus denotes the presence of nine carbon atoms in the compound. For instance, for alkanes, a nonane would be a nine carbon atom alkane, with non denoting the presence of nine carbon atoms, and <em>ane</em> being indicative that the compound it an alkane.

Dec is the prefix that is associated with the number nine, and thus denotes the presence of nine carbon atoms in the compound. For instance, for alkanes, a decane would be an alkane with ten carbon atoms, with dec denoting the presence of ten carbon atoms, and <em>ane</em> being indicative that the compound it an alkane.

<h3>Hope this helps!</h3>

4 0

3 years ago