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3 years ago

22 The compounds CH3OCH3 and CH3CH2OH have different functional groups. Therefore, these compounds have different

Chemistry

2 answers:

dangina [55]3 years ago

8 0

Answer:

(1) chemical properties

Explanation:

CH3OCH3 is dimethyl ether while CH3CH2OH is ethanol.

Both compounds have the same molecular formula i.e. C2H6O but different structural formula, therefore they are structural isomers. Since they have different functional groups i.e. ether (-OR) and alcohol (-OH), they will have different chemical properties.

The gram formula mass for both compounds is the same i.e. 46 g/mol. Based on their chemical formula, both will have the same percent compositions by mass as well as the same number of atoms per molecule.

Ivanshal [37]3 years ago

6 0

(1) chemical properties is the correct answer.
These two compounds are also called isomers.
Hope this helps~

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What is the meaning of the term polar as applied to chemical bonding?

kicyunya [14]

Polar covalent bonding is a type of chemical bond where a pair of electrons is unequally shared between two atoms.

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2 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

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2 years ago

Which of the following elements has the lowest first ionization energy?

Tanya [424]

Answer:

(Sc)

Explanation:

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3 years ago

Read 2 more answers

choose any bases that will selectively deprotonate the above acid, that is, any bases that will favor the formation of products.

Zanzabum

In order to deprotonate an acid, we must remove protons in order to achieve a more stable conjugate base. For this example, we can use the relationship between carboxylic acid and hydroxide.

Deprotonation is the removal of a proton from a specific type of acid in reaction to its coming into contact with a strong base. The compound formed from this reaction is known as the conjugate base of that acid. The opposite process is also possible and is when a proton is added to a special kind of base. This is a process referred to as protonation, which forms the conjugate acid of that base.

For the example we have chosen to give, the conjugate base is the carboxylate salt. This would be the compound formed by the deprotonated carboxylic acid. The base in question was strong enough to deprotonate the acid due to the greater stability offered as a conjugated base.

To learn more visit:

brainly.com/question/5613072?referrer=searchResults

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2 years ago

A sample of pure calcium fluoride with a mass of 15.0 g contains 7.70 g of calcium. how much calcium is contained in 45.0 g of c

kolbaska11 [484]

In a sample of pure calcium fluoride of mass 15.0 g, 7.70 g of calcium is present. First convert the mass into number of moles as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Molar mass of Ca is 40 g/mol, putting the values,

$n=\frac{7.70 g}{40 g/mol}=0.1925 mol$

Similarly, molar mass of $CaF_{2}$ is 78.07 g/mol thus, number of moles will be:

$n=\frac{15.0 g}{78.07 g/mol}=0.1921 mol$ .

Thus, 0.1921 mol of $CaF_{2}$ have 0.1925 mol of Ca, or 1 mole of $CaF_{2}$ will have approximately 1 mole of Ca.

Now, mass of Ca needs to be calculated in 45.0 g of $CaF_{2}$ . Converting mass into number of moles first,

$n=\frac{45.0 g}{78.07 g/mol}=0.5764 mol$

Thus, number of moles of Ca will also be 0.5764 mol, converting number of moles into mass,

$m=n\times M=0.5764 mol\times 40 g/mol=23.06 g$

Therefore, mass of Ca will be 23.06 g.

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3 years ago