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2 years ago

Someone please notice!! Explain the other parts such as the petal and receptacle do in the flower.

1 answer:

6 0

Answer: well the receptacle connect the stalk to the flower and to support the flower and keeps the flower in an elevated position so as to attract the insects

Explanation: I don’t know if this helped

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Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago

2 AICI3 + 3 Ca - 3 CaCl2 + 2 Al

Ivanshal [37]

50 grams of calcium

8 0

3 years ago

How many valence electrons would be used in drawing the Lewis Structure CF4?

Elis [28]

Answer:

Explanation:

You have 7 valance electrons in fluorine, and you have 4 of them. 7*4=28. Then, you have 4 valance electrons on carbon. So 28+4=32

3 0

3 years ago

Read 2 more answers

Which one <br> On my test

Karo-lina-s [1.5K]

Answer:

20 Hz

Explanation:

5 0

3 years ago

Radioactive carbon-14 has a half life of 5730 years. Suppose a peice of wood has a decay rate of 15 disintegrations per minute.

Tom [10]

Answer:

$\large \boxed{\text{2920 yr}}$

Explanation:

Two important formulas in radioactive decay are

$(1) \qquad t_{\frac{1}{2}} = \dfrac{\ln 2}{k}\\\\(2) \qquad \ln \left(\dfrac{N_{0}}{N}\right) = kt$

1. Calculate the decay constant k

$\begin{array}{rcl}t_{\frac{1}{2}} &=& \dfrac{\ln 2}{k}\\\\\text{1530 yr} &= &\dfrac{\ln 2}{k}\\\\k & = & \dfrac{\ln 2}{\text{1530 yr}}\\\\& = & 4.530 \times 10^{-4} \text{ yr}^{-1}\\\end{array}$

2. Calculate the time

$\begin{array}{rcl}\ln \left(\dfrac{N_{0}}{N}\right) &= &kt \\\\\ln \left(\dfrac{15}{4}\right) &= &4.530 \times 10^{-4} \text{ yr}^{-1}\times t \\\\\ln 3.75 &= &4.530 \times 10^{-4} \text{ yr}^{-1}\times t \\t &= &\dfrac{\ln 3.75}{4.530 \times 10^{-4} \text{ yr}^{-1}}\\\\& = & \textbf{2920 yr}\\\end{array}\\\text{It would take $\large \boxed{\textbf{2920 yr}}$ for the rate to decrease to 4 dis/min.}$

5 0

3 years ago