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3 years ago

How many compounds, of the ones listed below, have hydrogen bonding? ch3(ch2)2nh2 ch3(ch2)2nh(ch2)4ch3 (ch3ch2)2n(ch2)4ch3?

1 answer:

4 0

Answer:
Following two compounds have Hydrogen Bond Interactions;

1) CH₃(CH₂)₂NH₂ (Propan-1-amine)<span>

2) </span>CH₃(CH₂)₂NH(CH₂)₄CH₃ (N-propylpentan-1-amine)

Explanation:
Hydrogen Bond Interactions are formed between those molecules which has hydrogen atoms covalently bonded to most electronegative atoms like Fluorine, Oxygen and Nitrogen. This direct attachment of Hydrogen to electronegative atom makes it partial positive resulting in hydrogen bonding with neighbor's partial negative most electronegative atom. So, in above selected compounds it can be seen that both compounds contain hydrogen atoms directly attached to Nitrogen atoms, Therefore, allowing them to form Hydrogen Bonding Interactions.

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Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago

Descuss how to find the oldest paper in a stack of papers

deff fn [24]

It is by looking on the dates of the paper.

Hoped this helped.

~Bob Ross®

8 0

3 years ago

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A 2. 75-l container filled with co2 gas at 25°c and 225 kpa pressure springs a leak. When the container is re-sealed, the pressu

Serjik [45]

The number of moles of gas lost is 0.0213 mol. It can be solved with the help of Ideal gas law.

<h3>What is Ideal law ?</h3>

According to this law, "the volume of a given amount of gas is directly proportional to the number on moles of gas, directly proportional to the temperature and inversely proportional to the pressure. i.e.

PV = nRT.

Where,

p = pressure
V = volume (1.75 L = 1.75 x 10⁻³ m³)
T = absolute temperature
n = number of moles
R = gas constant, 8.314 J*(mol-K)

Therefore, the number of moles is

n = PV / RT

State 1 :

T₁ = (25⁰ C = 25+273 = 298 K)
p₁ = 225 kPa = 225 x 10³ N/m²

State 2 :

T₂ = 10 C = 283 K
p₂ = 185 kPa = 185 x 10³ N/m²

The loss in moles of gas from state 1 to state 2 is

Δn = V/R (P₁/T₁ - P₂/T₂ )

V/R = (1.75 x 10⁻³ m³)/(8.314 (N-m)/(mol-K) = 2.1049 x 10⁻⁴ (mol-m²-K)/N

p₁/T₁ = (225 x 10³)/298 = 755.0336 N/(m²-K)

p₂/T₂ = (185 x 10³)/283 = 653.7102 N/(m²-K)

Therefore,

Δn = (2.1049 x 10⁻⁴ (mol-m²-K)/N)*(755.0336 - 653.7102 N/(m²-K))

= 0.0213 mol

Hence, The number of moles of gas lost is 0.0213 mol.

Learn more about ideal gas here ;

https://brainly.in/question/641453

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3 0

2 years ago

The density of carbon in the form of diamond is 3.51 g/cm^3. if you have a small diamond with a volume of 0.0270 cm3 what is its

lora16 [44]

Hey there!:

density = 3.51 g/cm³

Volume = 0.0270 cm³

Therefore:

D = m / V

3.51 = m / 0.0270

m = 3.51 * 0.0270

m = 0.09477 g

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3 years ago

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How does the number of valence electrons for elements change across a period?

NeTakaya

D. The number increases and then decreases for noble gases

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3 years ago

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