Molar mass of (NaOH)

Which of the following best describes biogeochemical cycles? A. They cycle elements through the biotic and abiotic components of

Ad libitum [116K]

Answer:

A. They cycle elements through the biotic and abiotic components of an ecosystem.

Explanation:

Chemical elements such as nitrogen, carbon, phosphorus etc. moves between the living and non-living parts of an ecosystem in the biosphere. The cycle of these elements between these counterparts (abiotic and biotic) is referred to as the BIOGEOCHEMICAL CYCLE.

Firstly, the biotic components of an ecosystem refers to the living organisms such as plants, animals, microbes etc while the abiotic components are the non-living parts such as soil, water etc. Therefore, according to the description given in the options, option A describes what a BIOGEOCHEMICAL CYCLE best.

6 0

3 years ago

A 25.0-mL solution of 0.100 M CH3COOH

Natali5045456 [20]

Answer:

a) pH = 2,88

b) pH = 4,58

c) pH = 5,36

d) pH = 8,79

e) pH = 12,10

Explanation:

In a titration of a strong base (KOH) with a weak acid (CH₃COOH) the reaction is:

CH₃COOH + KOH → CH₃COOK + H₂O

a) Here you have just CH₃COOH, thus:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ where ka =1,74x10⁻⁵ and pka = 4,76

When this reaction is in equilibrium:

[CH₃COOH] = 0,100 -x

[CH₃COO⁻] = x

[H⁺] = x

Thus, equilibrium equation is:

1,74x10⁻⁵ = $\frac{[x][x] }{[0,100-x]}$

The equation you will obtain is:

x² + 1,74x10⁻⁵x - 1,74x10⁻⁶ = 0

Solving:

x = -0,0013278193 ⇒ No physical sense. There are not negative concentrations

x = 0,0013104193

As x = [H⁺] and pH = - log [H⁺]

pH = 2,88

b) Here, it is possible to use:

CH₃COOH + KOH → CH₃COOK + H₂O

With adition of 5,0 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = $0,025 L.\frac{0,100 mol}{L} =$ = 2,5x10⁻³ mol

KOH = $0,005 L.\frac{0,200 mol}{L} =$ = 1,0x10⁻³ mol

CH₃COOK = 0.

In equilibrium:

CH₃COOH = 2,5x10⁻³ mol - 1,0x10⁻³ mol = 1,5x10⁻³ mol

KOH = 0 mol

CH₃COOK = 1,0x10⁻³ mol

Now, you can use Henderson–Hasselbalch equation:

pH = 4,76 + log $\frac{1,0x10^{-3} }{1,5x10^{-3} }$

pH = 4,58

c) With adition of 10,0 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = $0,025 L.\frac{0,100 mol}{L} =$ = 2,5x10⁻³ mol

KOH = $0,010 L.\frac{0,200 mol}{L} =$ = 2,0x10⁻³ mol

CH₃COOK = 0.

In equilibrium:

CH₃COOH = 2,5x10⁻³ mol - 2,0x10⁻³ mol = 0,5x10⁻³ mol

KOH = 0 mol

CH₃COOK = 2,0x10⁻³ mol

Now, you can use Henderson–Hasselbalch equation:

pH = 4,76 + log $\frac{2,0x10^{-3} }{0,5x10^{-3} }$

pH = 5,36

d) With adition of 12,5 mL of 0,200M KOH solution the initial moles are:

CH₃COOH = $0,025 L.\frac{0,100 mol}{L} =$ = 2,5x10⁻³ mol

KOH = $0,0125 L.\frac{0,200 mol}{L} =$ = 2,5x10⁻³ mol

CH₃COOK = 0.

Here we have the equivalence point of the titration, thus, the equilibrium is:

CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ kb = kw/ka where kw is equilibrium constant of water = 1,0x10⁻¹⁴; kb = 5,75x10⁻¹⁰

Concentrations is equilibrium are:

[CH₃COOH] = x

[CH₃COO⁻] = 0,06667-x

[OH⁻] = x

Thus, equilibrium equation is:

5,75x10⁻¹⁰ = $\frac{[x][x] }{[0,06667-x]}$

The equation you will obtain is:

x² + 5,75x10⁻¹⁰x - 3,83x10⁻¹¹ = 0

Solving:

x = -0.000006188987⇒ No physical sense. There are not negative concentrations

x = 0.000006188

As x = [OH⁻] and pOH = - log [OH⁻]; pH = 14 - pOH

pOH = 5,21

pH = 8,79

e) The excess volume of KOH will determine pH:

With 12,5mL is equivalence point, the excess volume is 15,0 -12,5 = 2,5 mL

2,5x10⁻³ L × $\frac{0,200 mol}{1L}$ ÷ 0,040 L = 0,0125 = [OH⁻]

pOH = - log [OH⁻]; pH = 14 - pOH

pOH = 1,90

pH = 12,10

I hope it helps!

8 0

4 years ago

A generic element, Z, has two isotopes, 45Z and 47Z, and an average atomic mass of 45.36 amu. The natural abundances of the two

Dmitry_Shevchenko [17]

Answer:

46.96 amu

Explanation:

Isotopes are different kinds of same elements. The difference between two isotopes of the same element is the number of neutrons.

To get the relative atomic mass, we take into consideration the masses of the different isotopes. This is done by multiplying their abundances by their masses. They are then added together to get the relative atomic mass of the element.

Let the isotopic mass of 47Z be x

45.36 = [80/100 * 44.96] + [20/100 * x]

45.36 = 35.968 + 0.2x

0.2x = 45.36 - 35.968

0.2x = 9.392

x =9.392/0.2 = 46.96 amu

5 0

3 years ago

A compound X has the following percentage composition 66.7% carbon, 11.1% hydrogen and 22.2% oxygen .Calculate the empirical for

Digiron [165]

Explanation:

c. h. o

66.7%. 11.1%. 22.2%

____. ____. ____

12. 1. 16

1.558. 11.1. 1.39. (divide by the smallest)

1. 8. 1

empirical formula=ch8o

4 0

3 years ago

silver (I) nitrate reacts with nickel (II) chloride to produce silver (I) chloride and nickel (II) nitrate wright the balanced c

Naya [18.7K]

Answer: 2 AgNO3(aq) + NiCl2(aq) ⇒ 2 AgCl(s) + Ni(NO3)2 (aq)

Explanation:

4 0

3 years ago