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3 years ago

Science science science science

Chemistry

2 answers:

Paha777 [63]3 years ago

5 0

Answer:

Science

Explanation:

Science

In-s [12.5K]3 years ago

4 0

Answer:

Sorry

Explanation:

Sorry this is not chemistry but I always try to answer but this time I can't I am so so sorry

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How many chlorine atoms would be in 6.02 X 10^23 units of gold III chloride

Pepsi [2]

Answer:

The number of chlorine atoms present in $6.02 \times 10^{23}$ units of gold III chloride is $18.066 \times 10^{23}$

Explanation:

Formula of Gold (III) chloride: $AuCl_{3}$

Avogadro Number : Number of particles present in one mole of a substance.

${N_{0}} =6.022 \times 10^{23}$

Using,

$n(moles)=\frac{Given\ number\ of\ particles}{N_{0}}$

$n =\frac{6.02\times 10^{23}}{6.022\times 10^{23}}$

= 1 mole(0.9999 , nearly equal to 1 )

The given Gold III chloride sample is 1 mole in amount.

$6.022 \times 10^{23}$ = 1 mole of $AuCl_{3}$

In this Sample,

1 mole of $AuCl_{3}$ will give = 3 mole of Chlorine atoms

1 mole of Cl contain = $6.022 \times 10^{23}$

3 mole of Cl contain = $6.022 \times 10^{23}\times 3$

3 mole of Cl contain = $18.066 \times 10^{23}$

So,

The number of chlorine atoms present in $6.02 \times 10^{23}$ units of gold III chloride is $18.066 \times 10^{23}$

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