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3 years ago

A chunk of zinc is added to a solution of gold (III) nitrate to extract the gold. The reaction forms,

Chemistry

2 answers:

alina1380 [7]3 years ago

7 0

Answer:

it forms :

1. Gold ( Au )

2. Zinc nitrate ( Zn(NO3)2 )

Explanation:

When a chunk of zinc is added to a solution of gold (III) nitrate to extract the gold. The reaction forms Gold and Zinc nitrate .

it's a single displacement reaction,

here's the balanced equation for above reaction :

3 Zn + 2 Au(NO3)3 =》3 Zn(NO3)2 + 2 Au

Veseljchak [2.6K]3 years ago

4 0

Explanation:

Gold (AU)................

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Round to 4 significant figures.<br> 0.007062

lapo4ka [179]

Answer:

Hey there!

This is already rounded to four significant figures!

Zeroes after the decimal but before the 7 don't count, and 7, 0, 6, and 2 count as significant figures.

So, the answer would be 0.007062.

Let me know if this helps :)

8 0

3 years ago

The question is "How does the absorption and release of energy affect temperature change during a chemical reaction?" I need to

sergeinik [125]

If the heat is absorbed, then the temperature will increase as well.

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4 years ago

The advantage of asexual reproduction is.

Nataly_w [17]

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4 years ago

How many grams are in 1.76 x 10^23 atoms of iodine

Mariana [72]

Answer:

$\boxed {\boxed {\sf About \ 37.1 \ grams \ of \ iodine }}$

Explanation:

To convert from atoms to grams, you must first convert atoms to moles, then moles to grams.

1. Convert Atoms to Moles

To convert atoms to grams, Avogadro's number must be used.

$6.022*10^{23}$

This number tells us the number of particles (atoms, molecules, ions, etc.) in 1 mole. In this case, the particles are atoms of iodine.

$\frac{6.022*10^{23} \ atoms \ I }{1 \ mol \ I}$

Multiply the given number of atoms by Avogadro's number.

$1.76*10^{23} \ atoms \ I*\frac{6.022*10^{23} \ atoms \ I }{1 \ mol \ I}$

Flip the fraction so the atoms of iodine will cancel out.

$1.76*10^{23} \ atoms \ I*\frac{ 1 \ mol \ I}{6.022*10^{23} \ atoms \ I}$

$1.76*10^{23}* \frac{1 \ mol \ I}{6.022*10^{23} }$

Multiply so the problem condenses into 1 fraction.

$\frac{1.76*10^{23} \ mol \ I}{6.022*10^{23} }$

$0.2922617071 \ mol \ I$

2. Convert Moles to Grams

Now we must use the molar mass of iodine, which is found on the Periodic Table.

Iodine Molar Mass: 126.9045 g/mol

Use this mass as a fraction.

$\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}$

Multiply this fraction by the number of moles found above.

$0.2922617071 \ mol \ I*\frac{ 126.9045 \ g\ I }{ 1 \ mol \ I}$

Multiply. The moles of iodine will cancel.

$0.2922617071 *\frac{ 126.9045 \ g\ I }{ 1 }$

The 1 as a denominator is insignificant.

$0.2922617071 *{ 126.9045 \ g\ I }$

$37.08932581 \ g \ I$

3. Round

The original measurement of 1.76*10^23 has 3 significant figures (1, 7, and 6). Therefore we must round our answer to 3 sig figs. For this answer, that is the tenths place.

$37.08932581 \ g \ I$

The 8 in the hundredth place tells us to round the 0 up to a 1.

$\approx 37.1\ g \ I$

There is about <u>37.1 grams of iodine </u>in 1.76*10^23 atoms.

5 0

3 years ago

An oxide and carbonate containing oxygen

exis [7]

Answer:

example of carbonate containing oxygen :

CaCO3

( calcium carbonate having 3 atoms of oxygen )

example of oxide containing oxygen :

( carbon dioxide having 2 atoms of oxygen )

5 0

3 years ago