All categories

3 years ago

How many moles are in 9.25 x 10^58 atoms of copper

Chemistry

1 answer:

olga2289 [7]3 years ago

6 0

Hey there!

There are 6.022 x 10²³ atoms of copper in one mole of copper.

So, divide 9.25 x 10⁵⁸ by 6.022 x 10²³.

(9.25 x 10⁵⁸) ÷ (6.022 x 10²³) = 1.54

There are 1.54 moles of iron in 9.25 x 10⁵⁸ atoms of copper.

Hope this helps!

You might be interested in

Which substances will make a salt when combined?

babymother [125]

The answer is D. Fertilizer and vinegar

3 0

2 years ago

Read 2 more answers

I need help in this :(

alexandr402 [8]

Answer:

ok follow me twitter dm me for answer The Kid D is my twitter

Explanation:

6 0

2 years ago

Human skin color varies widely around the world, and children do not always exhibit the exact same coloring as their parents. Ba

makvit [3.9K]

D an example of polygenic inheritance

ps this is more biology than chemistry

6 0

2 years ago

What happens to the radius of non metal atom when it forms an ion?

MArishka [77]

Decrease because loss of electrons.

8 0

2 years ago

How many moles of FeCI3 are present in 345.0 g FeCI3

Sphinxa [80]

Answer:

$\boxed {\boxed {\sf About \ 2.127 \ moles \ of \ FeCl_3}}$

Explanation:

To convert from moles to grams, the molar mass must be used.

1. Find Molar Mass

The compound is iron (III) chloride: FeCl₃

First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).

Fe: 55.84 g/mol

Cl: 35.45 g/mol

There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.

FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol

This number tells us the grams of FeCl₃ in 1 mole.

2. Calculate Moles

Use the number as a ratio.

$\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}$

Multiply by the given number of grams, 345.0.

$345.0 \ g \ FeCl_3 *\frac{162.19 \ g \ FeCl_3}{1 \ mol \ FeCl_3}$

Flip the fraction so the grams of FeCl₃ will cancel.

$345.0 \ g \ FeCl_3 *\frac{1 \ mol \ FeCl_3}{162.19 \ g \ FeCl_3}$

$345.0 *\frac{1 \ mol \ FeCl_3}{162.19 }$

$\frac{345.0 \ mol \ FeCl_3}{162.19 }$

Divide.

$2.12713484 \ mol \ FeCl_3$

3. Round

The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.

For the answer we calculated, that is the thousandth place.

The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.

$\approx 2.127 \ mol \ FeCl_3$

There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.

6 0

2 years ago