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3 years ago

How many moles of Ca atoms are in 1 mol of CaSO4?

Chemistry

1 answer:

N76 [4]3 years ago

3 0

Hey there!

There is 1 mol of Ca atoms in 1 mol of CaSO₄

There is one atom of Ca in each molecule of CaSO₄, so in 6.022 x 10²³ molecules of CaSO⁴ there will be 6.022 x 10²³ atoms of Ca.

Hope this helps!

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The density of solid Ag is 10.5 g/cm3. How many atoms are present per cubic centimeter of Ag?

Sveta_85 [38]

Answer:

Unit cells which are present per cubic centimeter of Ag = $=1.45\times 10^{22}\ unit\ cells /cm^3$

Volume = $68.9\times 10^{-24}\ cm^3$

Edge length = $4.1\times 10^{-8}\ cm$

Explanation:

(a)

Given that:-

The density of the solid Ag = 10.5 g/cm³

Molar mass of silver = 107.8682 g/mol

So, Moles present per cm³ of Ag = $\frac{10.5\ g/cm^3}{107.8682\ g/mol}$ =0.0973 mol/cm³

Also, 1 mole = $6.023\times 10^{23}$ atoms.

So,

Atoms present per cm³ of Ag = $0.0973\ mol/cm^3\times 6.023\times 10^{23}\ atoms/mol=5.8\times 10^{22}\ atoms/cm^3$

Thus, answer = $5.8\times 10^{22}\ atoms/cm^3$

In FCC, the number of atoms in the unit cell = 4 unit cells

So,

Unit cells which are present per cubic centimeter of Ag = $\frac{5.8\times 10^{22}\ atoms/cm^3}{4}=1.45\times 10^{22}\ unit\ cells /cm^3$

Unit cells which are present per cubic centimeter of Ag = $=1.45\times 10^{22}\ unit\ cells /cm^3$

(b)

The reciprocal of the unit cell/cm³ is the volume of the unit cell.

So, $Volume=\frac{1}{1.45\times 10^{22}\ unit\ cells /cm^3}=68.9\times 10^{-24}\ cm^3$

Volume = $68.9\times 10^{-24}\ cm^3$

(c)

Also, Volume = ${(Edge\ length)}^3$

Thus, edge length = ${Volume}^{\frac{1}{3}}$ = $\left(68.9\times \:\:10^{-24}\right)^{\frac{1}{3}}\ cm=4.1\times 10^{-8}\ cm$

Edge length = $4.1\times 10^{-8}\ cm$

5 0

4 years ago

What is the mass of silver (107.87 g/mol) produced by the reaction of 3.00 moles of copper with 3.00 moles of silver nitrate?Cu

vladimir2022 [97]

Answer: 323.61 g of $Ag$ will be produced

Explanation:

The given balanced chemical reaction is :

$Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag$

According to stoichiometry :

2 moles of $AgNO_3$ require 1 mole of $Cu$

Thus 3.00 moles of $AgNO_3$ will require= $\frac{1}{2}\times 3.00=1.50moles$ of $Cu$

Thus $AgNO_3$ is the limiting reagent as it limits the formation of product.

As 2 moles of $AgNO_3$ give = 2 moles of $Ag$

Thus 3.00 moles of $AgNO_3$ give = $\frac{2}{2}\times 3.00=3.00moles$ of $Ag$

Mass of $Ag=moles\times {\text {Molar mass}}=3.00moles\times 107.87g/mol=323.61g$

Thus 323.61 g of $Ag$ will be produced from the given moles of both reactants.

3 0

3 years ago

When 25.0 grams of solid potassium hydroxide (KOH, molar mass = 56.1 g/mol) is dissolved in 100.0 grams of water, the solution i

Maru [420]

Answer:

They gave you the equation; Cp=,

just plug everything in! You’ve seen this; I have long ago, but we had different units. Sorry, but it’s right there! Go get it!

Explanation:

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4 years ago

A parachute on a racing dragster opens and changes the speed of the car from 85 m/s to 45 m/s in a period of 4.5 seconds. What i

astra-53 [7]

Answer:

I think the answer is -8.9 m/s2

8 0

4 years ago

When an isomer Y of molecular formula C4H9Br undergoes hydrolysis in aqueous alkali to form an alcohol C4H9OH, the rate of react

belka [17]

I would say D because the Br is bonded to a tertiary carbon (a carbon bonded to three other carbon atoms). This makes the bond more stable compared to if the Br was bonded to a primary or secondary carbon, and so it is unlikely for spontaneous substitution of Br for OH to occur. Hydrolysis occurs through reaction intermediates that make the making and breaking of C-OH and C-Br bonds respectively more favourable.

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3 years ago

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