All categories

3 years ago

How many moles are there in 3.80x10^22 molecules of CCI4?

Chemistry

1 answer:

Elena-2011 [213]3 years ago

3 0

Answer:

0.0646 mole.

Explanation:

The following data were obtained from the question:

Number of molecules = 3.89×10²² molecules

Number of mole =?

The number of mole present in 3.89×10²² molecules of CCl₄ can be obtained as follow:

From Avogadro's hypothesis:

6.02×10²³ molecules = 1 mole

Therefore,

3.89×10²² molecules = 3.89×10²² molecules × 1 mole / 6.02×10²³ molecules

3.89×10²² molecules = 0.0646 mole

Thus, 0.0646 mole is present in 3.89×10²² molecules of CCl₄.

You might be interested in

If the [H3O+] of a solution is 1x 10-8 mol/L the [OH-] is

Studentka2010 [4]

Answer:

(H30+)= 1x10^-6 M

Explanation:

Both pH and pOH have a relationship to belonging to the same aqueous solution: the expression of the Kwater (ionic product of the water Kw) is used:

1x 10-8 mol/L equals to1x10-8 M

(H3O+) x (OH-) = 1x10^-14

(H30+)x 1x 10^-8 =1x10^-14

(H30+)= 1x10^-14/1x 10^-8

(H30+)= 1x10^-6 M

6 0

3 years ago

If it requires 18.2 milliliters of 0.45 molar barium hydroxide to neutralize 38.5 milliliters of nitric acid, solve for the mola

Ghella [55]

Hi! :)

Ba(OH)2 + 2 HNO3 → Ba(NO3)2 + 2 H2O

(18.2 mL) x (0.45 M Ba(OH)2) x (2 mol HNO3 / 1 mol Ba(OH)2) / (38.5 mL HNO3) = 0.43 M HNO3

5 0

3 years ago

Read 2 more answers

A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate f

Alecsey [184]

Answer:

a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted = 0.326 g

e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g

g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaCl₂.2 H₂O in sample completely used up = 0.326 g

mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

6 0

2 years ago

Why are isotopes with short half-lives not useful for dating very old rocks?

erma4kov [3.2K]

Answer:

The radioactive isotopes within these brains typically record the ages of the grains-not the time when the sedimentary rock was formed, making it difficult to date using radiometric dating. The half-life of U-235 is 704 million years.

Explanation:

hope this helps but if i'm wrong then i'm sorry..

4 0

2 years ago

Read 2 more answers

Consider the reaction

spayn [35]

Answer:

1- 0.04 M/s.

2- 0.16 M/s.

Explanation:

For the reaction: 4PH₃ → P₄ + 6H₂.

The rate of the reaction = - d[PH₃]/4dt = d[P₄]/dt = d[H₂]/6dt.

where, - d[PH₃]/dt is the rate of PH₃ changing "rate of disappearance of PH₃".

d[P₄]/dt is rate of P₄ changing "rate of appearance of P₄".

d[H₂]/dt is the rate of H₂ changing "rate of formation of H₂" (d[H₂]/dt = 0.24 M/s).

(a) At what rate is P₄ changing?

∵ The rate of the reaction = d[P₄]/dt = d[H₂]/6dt.

∴ rate of P₄ changing = d[P₄]/dt = d[H₂]/6dt = (0.240 M/s)/(6.0) = 0.04 M/s.

(b) At what rate is PH₃ changing?

∵ The rate of the reaction = - d[PH₃]/4dt = d[H₂]/6dt.

∴ rate of PH₃ changing = - d[PH₃]/dt = 4(d[H₂]/6dt) = (4)(0.240 M/s)/(6.0) = 0.16 M/s.

7 0

3 years ago

Read 2 more answers