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3 years ago

Which type of fertilization is common with animals that live in the water?

Chemistry

2 answers:

Makovka662 [10]3 years ago

5 0

External fertilization in animals usually occurs in water or in damp areas in a process called spawning.

lutik1710 [3]3 years ago

4 0

The answer for this would be the first one external fertilization

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How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the so

expeople1 [14]

Answer:

$2.72$ grams

Explanation:

Complete question is

You are asked to pre pare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid

(C6H5COOH)

and any amount you need of sodium benzoate

(C6H5COONa)

How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

Solution

Given

pH of the buffer solution $= 4$

Concentration of C6H5COOH $= 0.02$ M

Volume of the buffer solution $= 1.50$ L

$K_a$ value for benzoic acid is $6.3 * 10^ {-5}$

Concentration of sodium benzoate

$pH = - log Ka + log \frac{C6H5COONa}{C6H5COOH}$

Substituting the given values we get

$log \frac{C6H5COONa}{C6H5COOH} = 4 + log (6.3 * 10^ {-5})\\log \frac{C6H5COONa}{C6H5COOH} = -0.20\\\frac{C6H5COONa}{C6H5COOH} = 10^{-0.2}\\{C6H5COONa} = 10^{-0.2} * 0.02\\{C6H5COONa} = 0.63 * 0.02\\{C6H5COONa} = 0.0126 M$

Number of moles in sodium benzoate

$= 0.0126 * 1.5 \\= 0.0189 Mol$

Mass of sodium benzoate

$0.0189 mol * 144.147 \frac{g}{mol} \\= 2.72 g$

3 0

3 years ago

What is the internal energy u of one mole of air on a very hot summer day (35∘c)? express your answer numerically in joules to t

umka21 [38]

- For solving this problem, we have to take in account the degree of freedom of air molecules.
- As, molecules of air have five degrees of freedom (three translational and two rotational)
- For each molecule, the kinetic energy of each degree of freedom = 1/2 KT
- So, K.E of 1 molecule for 5 degrees of freedom = 5/2KT
So, for molecules of air K.E = 5/2 KT = 5/2 x 1 x 1.38 x 10⁻²³ x 308 = 1062.6 x 10⁻²³ J
1 mole of air contains 6.022 x 10²³ molecules

K.E. of 1 mole = 1062.6 x 10⁻²³ x 6.022 x 10²³ = 6400 J

6 0

4 years ago

What volume of air contains 10.0g of oxygen gas at 273 k and 1.00 atm?

VARVARA [1.3K]

12.3g 02times1 mol 02/32.0 g 02=mol 02
.38mol/x=20.95/100
x=1.8mol air...

PV=nRT

101.3kPa times V =1.8 mol times 8.31[kPa *L]/{k*mol}*273k
V=40.3Lair

7 0

3 years ago

Read 2 more answers

What is it with the name Sara these days!? XD

Komok [63]

66 mph

Explanation:

Average speed is the rate of change of distance with time:

Average speed = $\frac{Distance }{time}$