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2 years ago

Because amines are basic, they can often accept a proton. draw the protonated structure of n-propylamine

1 answer:

3 0

Answer is: CH₃-CH₂-CH₂-NH₃⁺.
Amine group (-NH₂) has neutral charge, bet when accepts one proton (H⁺) has positive charge.
Amines are compounds that contain a basic nitrogen atom with a lone pair of electrons. Amines are derivatives of ammonia, where is one or more hydrogen atoms replaced by a substituent such as an alkyl or aryl group (in this compound propyl group).

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Explain how the models you developed show that when methane combines with oxygen to form carbon dioxide and water, no atoms are

Xelga [282]

Answer: the bonds in the methane and oxygen come apart, the atoms rearrange and then re-bond to form water and carbon dioxide

Explanation:^

3 0

2 years ago

(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o

Mrrafil [7]

Answer :

The correct answer is %IC = 10 % and bond is covalent bond with slight polarity.

Percent Ionic Character :

It is defined as percent of ionic character present in a polar covalent bond . The formula of % ionic character (%IC) is given as follows :

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100$

Where Xa = Electronegativity of A atom and Xb = Electronegativity of B atom

Given : Molecule is TiAl₃

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 ( From image shared )

Plug the value in above formula :

$Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100$

$Percent Ionic character = 1 - e^(^-^0^.^1^) * 100$

Value of e⁻¹ = 0.90

Percent ionic character = 1 - 0.90 * 100

Percent Ionic character = 10 %

Since the % IC is 10 % , which is very less comparatively , hence the bond is covalent and very less polar .

8 0

2 years ago

An object will sink in a liquid if the density of the object is greater than that of the liquid. the mass of a sphere is 9.83 g.

lord [1]

The volumeof this sphere must be less than 0.7228

5 0

2 years ago

A solution with a hydrogen ion concentration of 3.25 × 10-2 m is ________ and has a hydroxide concentration of _______

Romashka-Z-Leto [24]

To know the acidity of a solution, we calculate the pH value. The formula for pH is given as:

pH = - log [H+] where H+ must be in Molar

We are given that H+ = 3.25 × 10-2 M

Therefore the pH is:

pH = - log [3.25 × 10-2]

pH = 1.488

Since pH is way below 7, therefore the solution is acidic.

To find for the OH- concentration, we must remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,

[H+]*[OH-] = 10^-14 
[OH-] = 10^-14 / [H+]

[OH-] = 10^-14 / 3.25 × 10-2

[OH-] = 3.08 × 10-13 M

Answers:

Acidic

[OH-] = 3.08 × 10-13 M

6 0

2 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

1 year ago