Which statement describes a characteristic of a question that can be answered through scientific inquiry

A dry soil sample has a volume of 500 mL. After 400 mL of water are added to the soil, the soil becomes saturated with 150 mL of

Lerok [7]

The Answer is 50. hope this helps

4 0

3 years ago

What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa? Ka = 1.8×10-5

Westkost [7]

Answer:

a) We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

b) We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

c) We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

Explanation:

a) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = pK = -log(1.8*10^-5) = 4.74

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4.74 = 4.74 + log(A-/HA)

0 = log(A-/HA)

A-/HA = 1

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1

X =0.9

We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4.74 = 4.74 + log(0.9/0.9) = 4.74

b) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 4

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

4 = 4.74 + log(A-/HA)

-0.74 = log(A-/HA)

A-/HA = 0.182

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 0.182

X =0.277

We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4 = 4.74 + log(0.277/1.523)

c) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 5

Step 2:

Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

5 = 4.74 + log(A-/HA)

0.26 = log(A-/HA)

A-/HA = 1.82

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 1.82

X =1.16

We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

5 = 4.74 + log(1.16/0.64) = 5

3 0

4 years ago

The volume of gas is 252 mL at 340.0 kPa pressure. What will the volume be when the pressure is reduced to 50.0 kPa? Temperature

MrRissso [65]

Answer:

Clearly volume will INCREASE almost sevenfold, as we would expect if we reduce the pressure. Marilia E.K. The final volume of this gas will be 1.7L.

4 0

3 years ago

How large is the observable universe

Norma-Jean [14]

Here goes the question with no exact answer, lol <3

Scientists tell us that the observable universe is about 90 light years but then I wonder how they calculated that because the universe is immense!

I'm sorry but I can't go deeper into explaining as that's a tough question.

7 0

3 years ago

All of the following reactions can be described as displacement reactions except:____________.

Lady_Fox [76]

Answer:

b

Explanation:

The reaction that is not a displacement reaction from all the options is $C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}$

In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.

Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.

Correct option = b.

8 0

3 years ago