How many molecules are present in a 0.39 mole sample of oxygen gas?​

Sagot :

Answer:

Hello!

How many atoms of cobalt are in a 0.39 mole sample of Co(C2H3O2)3?

* Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10²³ molecules.

1 mol ------------------ 6.022*10²³ molecules

0.39 mol --------------------------- y molecules

\dfrac{1}{0.39} = \dfrac{6.022*10^{23}}{y}

0.39

1

=

y

6.022∗10

23

multiply the means by the extremes

1*y = 0.39*6.022*10^{23}1∗y=0.39∗6.022∗10

23

\boxed{y = 2.34858*10^{23}\:molecules\:of\:Co(C_2H_3O_2)_3}

y=2.34858∗10

23

moleculesofCo(C

2

H

3

O

2

)

3

* We know that an molecule is formed by one cobalt atom, so we have:

1 molecule of Co(C2H3O2)3 --------------------------------- 1 atom of Co

2.34858*10²³ molecules of Co(C2H3O2)3 -------------- y atom of Co

\dfrac{1}{2.34858*10^{23}} = \dfrac{1}{y}

2.34858∗10

23

1

=

y

1

multiply the means by the extremes

1*y = 1*2.34858*10^{23}1∗y=1∗2.34858∗10

23

y = 2.34858*10^{23}y=2.34858∗10 Hello!

How many atoms of cobalt are in a 0.39 mole sample of Co(C2H3O2)3?

* Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10²³ molecules.

1 mol ------------------ 6.022*10²³ molecules

0.39 mol --------------------------- y molecules

\dfrac{1}{0.39} = \dfrac{6.022*10^{23}}{y}

0.39

1

=

y

6.022∗10

23

multiply the means by the extremes

1*y = 0.39*6.022*10^{23}1∗y=0.39∗6.022∗10

23

\boxed{y = 2.34858*10^{23}\:molecules\:of\:Co(C_2H_3O_2)_3}

y=2.34858∗10

23

moleculesofCo(C

2

H

3

O

2

)

3

* We know that an molecule is formed by one cobalt atom, so we have:

1 molecule of Co(C2H3O2)3 --------------------------------- 1 atom of Co

2.34858*10²³ molecules of Co(C2H3O2)3 -------------- y atom of Co

\dfrac{1}{2.34858*10^{23}} = \dfrac{1}{y}

2.34858∗10

23

1

=

y

1

multiply the means by the extremes

1*y = 1*2.34858*10^{23}1∗y=1∗2.34858∗10

23

y = 2.34858*10^{23}y=2.34858∗10

23

\boxed{\boxed{y \approx 2.35*10^{23}\:Atoms\:of\:Cobalt}}\end{array}}\qquad\checkmark

Answer:

Approximately 2.35*10²³ atoms of cobalt123

23

\boxed{\boxed{y \approx 2.35*10^{23}\:Atoms\:of\:Cobalt}}\end{array}}\qquad\checkmark

Answer:

Approximately 2.35*10²³ atoms of cobalt

Go Educations: Other Questions