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prohojiy [21]
3 years ago
8

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

Chemistry
1 answer:
Aloiza [94]3 years ago
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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The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
Write the balanced equation. Then calculate the volume of 0.65 M HCl required to completely neutralize 400.0 ml of 0.88 M KOH.
Misha Larkins [42]
Hello!

The balanced equation for the neutralization of KOH is the following:

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

To calculate the volume of HCl required, we can apply the following equation:

 moles HCl = moles KOH \\  \\ cHCl*vHCl=cKOH*vKOH \\  \\ vHCl= \frac{cKOH*vKOH}{cHCl}= \frac{400 mL*0,88M}{0,65M}=  541,54mL

So, the required volume of HCl is 541,54 mL

Have a nice day!
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