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4 years ago

9

Differentiate between gamopetalous and polypetalous

2 answers:

Dominik [7]4 years ago

6 0

Answer:

As adjectives the difference between polypetalous and gamopetalous. is that polypetalous is (botany) having a corolla composed of distinct, separable petals while gamopetalous is (botany) having petals wholly or partially fused such that the corolla takes the form of a tube.

Explanation:

Art [367]4 years ago

3 0

Answer:

the difference between polypetalous and gamopetalous, is that polypetalous is (botany) having a corolla composed of distinct, separable petals while gamopetalous is (botany) having petals wholly or partially fused such that the corolla takes the form of a tube.Explanation:

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3 years ago

Be sure to answer all parts.

Aleks04 [339]

Answer:

pH = 3.215

Explanation:

From the given information;

Using the equation for the dilution of a stock solution:

Since moles of C5H5N = moles of HNO3

Then:

$M_{C_5H_5N}\times V_{C_5H_5N}= M_{HNO_3}\times V_{HNO_3}$

$0.0750 M \times 2.65 L = 0.407 M \times V_{HNO_3}$

$V_{HNO_3}= \dfrac{0.0750 M \times 2.65 L}{0.407 M}$

$V_{HNO_3}=488.33 \ mL$

The reaction between C5H5NH and H2O is as follows:

$C_5H_5N^+H + H_2O$ ⇄ $C_5H_5N + H_3O^+$

$Molarity \ of \ C_5H_5N^+H = \dfrac{moles}{addition \ of\ the\ total\ volume}$

$Molarity \ of \ C_5H_5N^+H = \dfrac{0.0750 \ M \times 2.65 \ L}{2.65 \ L + 0.48833 \ L}$

$Molarity \ of \ C_5H_5N^+H = \dfrac{0.19875 \ ML}{3.13833 L }$

$= \ 0.06333\ M$

Now, the next step is to draw out the I.C.E table.

$C_5H_5N^+H + H_2O$ ⇄ $C_5H_5N + H_3O^+$

I 0.06333 0 0

C - x x x

E 0.06333 -x x x

$K_a = \dfrac{10^{-14}}{1.7 \times 10^{-19}} \\ \\ K_a = 5.8824 \times 10^{-6}$

$K_a = \dfrac{[C_5H_5N][H_3O^+] }{[C_5H_5N^+H] } \\ \\ 5.8824 \times 10^{-6} = \dfrac{x^2}{0.06333 - x}$

Assuming x < 0.06333

$x^2 = 5.8824 \times 10^{-6} \times 0.06333$

Then

$x^2 = 3.72532392 \times 10^{-7} \\ \\ x= \sqrt{3.72532392 \times 10^{-7}} \\ \\ x = 6.1035 \times 10^{-4} \ M$

$[H_3O^+] = x = 6.1035 \times 10^{-4} \ M \\ \\ pH = -log (6.1035 \times 10^{-4}) \\ \\ \mathbf{\\ pH = 3.215}$

6 0

3 years ago