What is the mass of butane gas, C4H10, that can be held in a 3.00 L container at STP?

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium

lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0

3 years ago

What mass of copper is required to replace silver from 4.00g of silver nitrate dissolved in water?

larisa [96]

The balanced chemical reaction is written as:

Cu +2AgNO3 → Cu(NO3)2 + 2Ag

We are given the amount of silver nitrate to be used for the reaction. This value will be the starting point of our calculations. It is as follows:

4.00 g AgNO3 ( 1 mol / 169.87 g ) ( 1 mol Cu / 2 mol AgNO3 ) ( 63.456 g / 1 mol ) = 0.747 g Cu

3 0

3 years ago

What are three possible blood type alleles?

Keith_Richards [23]

Answer:

Three possible blood type alleles are Iᴬ, Iᴮ and i

Explanation:

Iᴬ, Iᴮ and i are three possible blood type alleles.

Iᴬ and Iᴮ are known as co-dominant, and The i allele is recessive.

Thus, Three possible blood type alleles are Iᴬ, Iᴮ and i

-TheUnknownScientist

8 0

2 years ago

Read 2 more answers

The Nutrition Facts label for crackers states that one serving contains 19 g of carbohydrate, 4 g of fat, and 2 g of protein. Wh

sp2606 [1]

Answer: The amount of kilocalories contained in the given serving of crackes is 0.120 kCal

Explanation:

As per the USDA:

Carbohydrates provide 4 calories per gram, protein provides 4 calories per gram, and fat provides 9 calories per gram

We are given:

Mass of fat in the crackers = 4 g

Mass of carbohydrates in the crackers = 19 g

Mass of protein in the crackers = 2 g

Conversion factor used: 1 kCal = 1000 Cal

Applying unitary method:

For Fat:

1 gram of fat provides 9 calories

So, 4 gram of fat will provide = $\frac{9}{1}\times 4=36Cal=0.036kCal$

For Carbohydrates:

1 gram of carbohydrates provides 4 calories

So, 19 gram of carbohydrates will provide = $\frac{4}{1}\times 19=76Cal=0.076kCal$

For Proteins:

1 gram of proteins provides 4 calories

So, 2 gram of fat will provide = $\frac{4}{1}\times 2=8Cal=0.008kCal$

Total kilocalories per serving = [0.036 + 0.076 + 0.008] kCal = 0.120 kCal

Hence, the amount of kilocalories contained in the given serving of crackes is 0.120 kCal

8 0

3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

2 years ago