All categories

3 years ago

Which of these choices is a benefit of international trade?

Chemistry

1 answer:

Inessa [10]3 years ago

4 0

Fewer option for consumers

You might be interested in

Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic aci

Anton [14]

Answer:

15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.

Explanation:

To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid

4.25 = 3.75 + log [A⁻] / [HA]

0.5 = log [A⁻] / [HA]

3.162 = [A⁻] / [HA] (1)

As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:

0.10M * 20x10⁻³L =

2x10⁻³moles = [A⁻] + [HA] (2)

Replacing (2) in (1):

3.162 = 2x10⁻³moles - [HA] / [HA]

3.162 [HA] = 2x10⁻³moles - [HA]

4.162[HA] = 2x10⁻³moles

[HA] = 4.805x10⁻⁴ moles

[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles

That means, to create the buffer you must add:

[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =

<h3>15.2mL of the 0.10M sodium formate solution</h3>

[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =

<h3>4.8mL of the 0.10M formic acid solution</h3>

4 0

3 years ago

If boron atoms have 5 electrons, how many electrons are in the outer electron shell of a boron atom?

ANEK [815]

Answer:

Explanation:

The first election shell can only hold 2 electrons, but the next one can hold up to 8

3 0

2 years ago

Valence electrons Fill in the blanks

IgorC [24]

Answer:

Valence electrons or outer electrons are most important as they participate in bonding. The octet rule states that atoms gain, lose, or share valence electrons to have filled energy levels.. this gives atoms a stable configuration like that of the nearest noble gas.

6 0

3 years ago

Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after

Alenkinab [10]

Answer:

For a: The number of moles of air present in the RV is 0.047 moles

For b: The number of molecules of gas is $2.83\times 10^{22}$

Explanation:

For a:

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the air = $37^oC=[37+273]K=310K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of air = ?

Putting values in above equation, we get:

$1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol$

Hence, the number of moles of air present in the RV is 0.047 moles

For b:

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.047 moles of air will contain $(0.047\times 6.022\times 10^{23})=2.83\times 10^{22}$ number of gas molecules.

Hence, the number of molecules of gas is $2.83\times 10^{22}$

7 0

3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

3 years ago