Ask question

Ask question

All categories

2 years ago

12

Where do producers get matter and energy to live and grow ?

1 answer:

bixtya [17]2 years ago

8 0

Answer:c

Explanation:

You might be interested in

The following equation is an example of a ______________ reaction. 2 NaCl + F2 → 2 NaF + Cl2 ???

olga55 [171]

Answer:

double replacement is the answer

3 0

2 years ago

If you were going to graphically determine the enthalpy of vaporization, δhvap, for this liquid, what points would you plot

Klio2033 [76]

It depends on what type of graph you have. The easiest would be using a H-T diagram. Enthalpy of vaporization is the physical change from liquid to vapor. It occurs at a constant pressure and a constant temperature. As shown in the picture, 1 point is drawn on the subcooled liquid, and another point of the saturated vapor isothermal line. Now, the difference between those two points is the value for the enthalpy of vaporization of water.

8 0

3 years ago

To study earths interior, geologist often rely on indirect methods, such as evidence from fossils.

igomit [66]

Answer: false

Explanation:

It is false that to study Earth's interior, geologists often rely on indirect methods, such as evidence from fossils. They rely on seismic wave.

6 0

2 years ago

Read 2 more answers

Quick! <br>I need to know <br>about how<br>old is the <br>earth!?

Salsk061 [2.6K]

Answer:

4.543 billion years old

3 0

3 years ago

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

2 years ago