Does absorbance increase or decrease as the solution concentration of the absorbing substance increases?

pashok25 [27]

The role of a decomposer is to decompose or break down dead matter in the environment. Plants make their own food by the process of photosynthesis and also produce (hence the name) food for other consumers. Without producers an ecosystem could not sustain itself.

6 0

3 years ago

Need help with 22 and 24

choli [55]

Answer:

22:

Formular:

$atomic \: mass = \frac{ \sum(isotopic \: mass \times \%abundance)}{100} \\$

substitute:

$atomic \: mass = \frac{(23.985 \times 78.70) + (24.986 \times 10.13) + (25.983 \times 11.17)}{100} \\ \\ = \frac{(1887.620) + (253.108) + (290.230)}{100} \\ \\ = \frac{2430.958}{100} \\ \\ { \boxed{ \boxed{average \: atomic \: mass = 24.3 \: amu}}}$

23:

Same element is represented by same number of protons. 

Answer:

6 protons. 6 protons

7 neutrons. 8 neutrons

6 electrons. 6 electrons

Note: Atoms with same proton number but different mass number are called isotopes

5 0

3 years ago

Read 2 more answers

Can you treat a sulfuric acid burn with sodium hydroxide

Alina [70]

Answer:

I think yes

Explanation:

sorry i havent done these type a questions in a while

3 0

2 years ago

A student determines that she used 0.0665 mol of sodium hydroxide (NaOH) to completely titrate 25.00 mL of sulfuric acid solutio

AysviL [449]

To determine the concentration of one solution which is specifically basic or acidic solution through taking advantage on its points of equivalence, titration analysis is done.

Let us determine the reaction for the titration below:

2NaOH +2H2SO4 = Na2SO4 +2H2O

So,

0.0665 mol NaOH (2 mol H2SO4/ 2mol NaOH) / .025 L solution

= 2.62 M H2SO4

The answer is the fourth option:

2.62 M

7 0

3 years ago

Read 2 more answers

If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a

vodomira [7]

Answer:

pH = 4.543

Explanation:

CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ C CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ C CH3CH2COOH = 0.048 M

⇒ C NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

3 0

3 years ago