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2 years ago

How many atoms are in a sample of 0.75 moles of carbon atoms

Chemistry

2 answers:

vova2212 [387]2 years ago

8 0

Answer: 4.5165 *10^-23 moles Carbon

Explanation:

There are 6.022*10^23 atoms in a mole.

Therefore, there would be 6.022*10^-23*0.75 atoms in 0.75 moles of Carbon.

= 4.5165 *10^-23 moles Carbon

ryzh [129]2 years ago

3 0

Answer: 4.5165 *10^-23 moles Carbon

Explanation:

There are 6.022*10^23 atoms in a mole.

Therefore, there would be 6.022*10^-23*0.75 atoms in 0.75 moles of Carbon.

= 4.5165 *10^-23 moles Carbon

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1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

A sealed 1.0L flask is filled with 0.500 mols of I_2 and 0.500 mols of Br_2. When the container achieves equilibrium the equilib

DochEvi [55]

Answer:

[IBr] = 0.049 M.

Explanation:

Hello there!

In this case, according to the balanced chemical reaction:

$I_2+Br_2\rightarrow 2IBr$

It is possible to set up the following equilibrium expression:

$K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110$

Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of $x$ (reaction extent) would be:

$0.0110=\frac{(2x)^2}{(0.50-x)^2}$

Which can be solved for $x$ to obtain two possible results:

$x_1=-0.0277M\\\\x_2=0.0245M$

Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:

$[IBr]=2x=2*0.0249M=0.049M$

Regards!

7 0

2 years ago

Student

s2008m [1.1K]

Does it is a complete qwestion? It

is not clearef

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3 years ago

what object experiences a frequency equal to its natural frequency _______ is said to have occurred. 1. resonance 2. frequency 3

Talja [164]

Amplitude is the awnser glad I could help you

8 0

2 years ago

An object with a mass of 5.6 g raises the level of water in a graduated cylinder from 25.1 mL to 33.9 mL. What is the density of

kondor19780726 [428]

Answer:

0.6364 g/cm^3

Explanation:

Density = mass/volume

Where mass = 5.6g and...

Volume = (33.9 - 25.1) = 8.8ml

Where 1ml = 1 cm^3

Density = 5.6/8.8 = 0.6364 g/cm^3

6 0

3 years ago

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