Which of these actions would the cerebellum be in charge of?

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

3 years ago

Two different bromide solutions are mixed with each other: Solution 1 is an aqueous solution of 4.85 g aluminum bromidein 150. m

erma4kov [3.2K]

Answer:

M=0.380 M.

Explanation:

Hello there!

In this case, given those two solutions of aluminum bromide and zinc bromide, it is firstly necessary to compute the moles of bromide ions in each solution as shown below:

$n_{Br^-}^{in\ AlBr_3}=4.85 gAlBr_3*\frac{1molAlBr_3}{266.69gAlBr_3}*\frac{3molBr^-}{1molAlBr_3} =0.05456molBr^-\\\\n_{Br^-}^{in\ ZnBr_2}=7.75gZnBr_2*\frac{1molZnBr_2}{225.22gZnBr_2}*\frac{2molBr^-}{1molZnBr_2} =0.06882molBr^-$