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3 years ago

13

How many moles of gold are equivalent to 1.204 × 1024 atoms?

2 answers:

Zinaida [17]3 years ago

8 0

How many moles of gold are equivalent to 1.204 × 1024 atoms?

0.2

0.5

2 CORRECT ANSWER

5

C) 2 Is the correct answer, I took the test and it was correct.

LUCKY_DIMON [66]3 years ago

7 0

Explanation:

It is known that the value of Avogadro's number is $6.022 \times 10^{23}$ .

Therefore, calculate moles of gold atoms as follows.

Moles of gold atom = $\frac{given number of atoms}{Avogadro's number}$

= $\frac{1.204 \times 10^{24}}{6.022 \times 10^{23}mol^{-1}}$

= $0.199 \times 10$ mol

= 1.99 mol

Thus, we can conclude that 1.99 mol of gold atoms are equivalent to $1.204 \times 10^{24}$ atoms.

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help plz Which list is in order from smaller to larger sediment particles? A. sand, silt, clay B. clay, silt, sand C. silt, sand

MariettaO [177]

Answer : The correct option is B (clay, slit, sand).

Explanation :

Soil particles are of three types : Sand, Slit and clay. Most of the soils particles are made up of a combination of sand, slit and clay.

Particle size range of the sand = 2.00 -0.05 mm

Particle size range of the slit = 0.05 -0.002 mm

Particle size range of the sand = less than 0.002 mm

Sand is the largest soil particle, clay is the smallest soil particle and slit particle is present in between the sand & clay particle.

Therefore, the order from smaller to larger sediment particles is

Clay > Slit > Sand

5 0

4 years ago

If you had an aqueous mixture that contained Ag+ , K+ , and Pb2+ cations, how many different solids could precipitate if a chlor

SCORPION-xisa [38]

Answer:

Two, KCl and PbCl₂.

Explanation:

Hello!

In this case, since the addition of chloride ions promote the following three ionic reactions:

$Ag^+(aq)+Cl^-(aq)\rightleftharpoons AgCl(s)\\\\K^+(aq)+Cl^-(aq)\rightleftharpoons KCl(aq)\\\\Pb^{2+}(aq)+Cl^-(aq)\rightleftharpoons PbCl_2(s)$

We can infer that both silver chloride and lead (II) chloride are precipitated products as their Ksp are 6.56x10⁻⁴ and 1.59x10⁻⁵ respectively, which means they are merely soluble in water.

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3 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

4 years ago

Calculate the number of milliliters of 0.440 M KOH required to precipitate all of the Fe2+ ions in 187 mL of 0.692 M FeSO4 solut

EleoNora [17]

Answer:

588.2 mL

Explanation:

FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)

First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):

moles = molarity * volume

187 mL * 0.692 M = 129.404 mmol Fe⁺²

Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:

129.404 mmol Fe⁺² * $\frac{2mmolKOH}{1mmolFeSO_4}$ = 258.808 mmol KOH

Finally we<u> calculate the required volume of KOH solution</u>, using <em>the given concentration and the calculated moles</em>:

volume = moles / molarity

258.808 mmol KOH / 0.440 M = 588.2 mL

6 0

3 years ago

Why is it important to use low flame when evaporating water from a recovered filtrate?

goldfiish [28.3K]

It is important to use low flame when evaporating water from a recovered filtrate because then the water and filtrate will not spatter and the filtrate can also be recovered after evaporating water.

If flame is not low then water as well as got spatter so it is important to use low flame so that the water and filtrate will not spatter.

4 0

4 years ago