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3 years ago

PLEASE HELPP!!!!

Chemistry

1 answer:

Alenkinab [10]3 years ago

3 0

Answer:

Na the valency electrons is 1

The ion charge is positive 1

K the valency electrons is 1

The ion charge is positive 1

Explanation:

Mg the valency electron is 2 The ion charge is positive 2

Ca the valency electron is 2 The ion charge is positive 2

Al the valency electron is 3 The ion charge is positive 3

O the valency electron is 2 The ion charge

is negative 2

F the valency electron is 1 The ion charge is negative 1

Cl the valency electron is 1 The ion charge

is negative 1

N the valency electron is 3 The ion charge is negative 3

S the valency electron is 2 The ion charge is

negative 2

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Erbium (Er) is element 68 on the periodic table. A sample contains 5.04×1023 atoms of Er. Calculate the amount of Er.

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Answer:

139.98 g to nearest hundredth.

Explanation:

Using Avogadro's Number:

One mole (167.26 g) of Erbium equates to 6.022141 * 10^23 atoms.

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3 years ago

(I)how many atoms are present in 7g of lithium?

ICE Princess25 [194]

Answer :

(i) The number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

(ii) The number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

(iii) The number of moles of $F_2$ is, 1 mole

The number of moles of $CO_2$ is, 0.5 mole

The number of moles of $OH^-$ is, 1 mole

Explanation :

Part (i) :

First we have to calculate the moles of lithium.

$\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}$

Molar mass of Li = 6.94 g/mole

$\text{Moles of }Li=\frac{7g}{6.94g/mol}=1.008mole$

Now we have to calculate the number of atoms present.

As, 1 mole of lithium contains $6.022\times 10^{23}$ number of atoms

So, 1.008 mole of lithium contains $1.008\times 6.022\times 10^{23}=6.07\times 10^{23}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $6.07\times 10^{23}$

Part (ii) :

First we have to calculate the moles of carbon.

$\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}$

Molar mass of C = 12 g/mole

$\text{Moles of }C=\frac{24g}{12g/mol}=2mole$

Now we have to calculate the number of atoms present.

As, 1 mole of carbon contains $6.022\times 10^{23}$ number of atoms

So, 2 mole of carbon contains $2\times 6.022\times 10^{23}=1.204\times 10^{24}$ number of atoms

Thus, the number of atoms present in 7 g of lithium are, $1.204\times 10^{24}$

Part (iii) :

To calculate the moles of $F_2$ :

$\text{Moles of }F_2=\frac{\text{Mass of }F_2}{\text{Molar mass of }F_2}$

Molar mass of $F_2$ = 38 g/mole

$\text{Moles of }F_2=\frac{19g}{19g/mol}=1mole$

Thus, the number of moles of $F_2$ is, 1 mole

To calculate the moles of $CO_2$ :

$\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}$

Molar mass of $CO_2$ = 44 g/mole

$\text{Moles of }CO_2=\frac{22g}{44g/mol}=0.5mole$

Thus, the number of moles of $CO_2$ is, 0.5 mole

To calculate the moles of $OH^-$ ions :

$\text{Moles of }OH^-=\frac{\text{Mass of }OH^-}{\text{Molar mass of }OH^-}$

Molar mass of $OH^-$ = 17 g/mole

$\text{Moles of }OH^-=\frac{17g}{17g/mol}=1mole$

Thus, the number of moles of $OH^-$ is, 1 mole

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