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Alenkasestr [34]

3 years ago

15

Why would sand and water be an example of a heterogeneous mixture? Select all that apply. Chemical bonds are formed between the

sand and water. *The mixture does not have a uniform appearance. *The mixture does not maintain a uniform composition. The sand stays suspended in the water.

2 answers:

drek231 [11]3 years ago

7 0

The mixture does not maintain uniform composition and sand stays suspended in water. And no, there no chemical bond between sand and water.

Sliva [168]3 years ago

6 0

<span>The mixture does not have a uniform appearance. </span><span>
The sand stays suspended in the water. (The sand sinks actually but I think by saying suspended, they mean it does not dissolve in water)</span>

You might be interested in

Describe how the metal probably increases the reaction rate, identify whether this is an example of homogeneous or heterogeneous

Lisa [10]

Answer:

The metal probably increases reaction rate by either holding reactant molecules in the correct orientation to react or by weakening or breaking bonds in reactant molecules to make them more reactive.

This is an example of heterogeneous catalysis.

It is heterogeneous catalysis because the catalyst is a solid and the reactants are gases. In heterogeneous catalysis, the catalyst is in a different phase than the reactants

Explanation:

got it right :)

8 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

Question 38 (1 point)

Fiesta28 [93]

Answer: Protons because they have a positive charge.

Explanation:

8 0

3 years ago

A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.

yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH since after reaction we will have both the conjugate base lactate anion and unreacted weak lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA = (20.0 mL x 1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴ ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

HA + NaOH ⇒ A⁻ + H₂O

before rxn 0.002 0.0005 0

after rxn 0.002-0.0005 0 0.0005

0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

HA + NaOH ⇒ A⁻ + H₂O

before rxn 0.002 0.001 0

after rxn 0.002-0.001 0 0.001

0.001

pH = 3.85 + log(0.001/0.001) = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

HA + NaOH ⇒ A⁻ + H₂O

before rxn 0.002 0.002 0

after rxn 0 0 0.002

We are at the neutralization point and we do not have a buffer anymore, instead we just have a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻] = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log ( 1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

HA + NaOH ⇒ A⁻ + H₂O

before rxn 0.002 0.0025 0

after rxn 0 0.0005 0.0005

Now here what we have is the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0

3 years ago

What is the edge length of a face-centered cubic unit cell that is made of of atoms, each with a radius of 154 pm

gladu [14]

Answer:

The edge length of a face-centered cubic unit cell is 435.6 pm.

Explanation:

In a face-centered cubic unit cell, each of the eight corners is occupied by one atom and each of the six faces is occupied by a single atom.

Hence, the number of atoms in an FCC unit cell is:

$8*\frac{1}{8} + 6*\frac{1}{2} = 4 atoms$

In a face-centered cubic unit cell, to find the edge length we need to use Pythagorean Theorem:

$a^{2} + a^{2} = (4R)^{2}$ (1)

Where:

a: is the edge length

R: is the radius of each atom = 154 pm

By solving equation (1) for "a" we have:

$a = 2R\sqrt{2} = 2*154 pm*\sqrt{2} = 435.6 pm$

Therefore, the edge length of a face-centered cubic unit cell is 435.6 pm.

I hope it helps you!

7 0

4 years ago