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2 years ago

3.51e23 atoms Fe= g Fe

Chemistry

1 answer:

lubasha [3.4K]2 years ago

7 0

Answer:

Mass of Fe = 32.55 g

Explanation:

Molar Mass of Fe atom = 55.845 g/mol

It means,

1 mole of Fe contains = 55.845 g ..............(1)

From mole concept,

1 mole of Fe contain = $6.022\times 10^{23}$ atom ..........(2)

Replace 1 mole in equation (2) by 55.845 g , it become :

55.845 g of Fe contain = $6.022\times 10^{23}$ atoms

Reverse the equation now

$6.022\times 10^{23}$ atoms of Fe contain =55.845 g

1 atom of Fe contain = $\frac{55.845}{6.022\times 10^{23}}$ g

$3.51\times 10^{23}$ atom contain =

$\frac{55.845}{6.022\times 10^{23}}\times 3.51\times 10^{23}$

on calculation,

$3.51\times 10^{23}$ atom contain = 32.55 g of Fe

Mass of Fe = 32.55 g

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2. A compound with the following composition by mass: 24.0% C, 7.0% H, 38.0% F, and 31.0% P. what is the empirical formula

Svetach [21]

Answer:

C₂H₇F₂P

Explanation:

Given parameters:

Composition by mass:

C = 24%

H = 7%

F = 38%

P = 31%

Unknown:

Empirical formula of compound;

Solution :

The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;

C H F P

% composition

by mass 24 7 38 31

Molar mass 12 1 19 31

Number of

moles 24/12 7/1 38/19 31/31

2 7 2 1

Dividing

by the

smallest 2/1 7/1 2/1 1/1

2 7 2 1

Empirical formula C₂H₇F₂P

5 0

2 years ago

Which of the following represents a propagation step in the monochlorination of methylene chloride (CH2Cl2)?a. CHCl3 + Cl. Right

svetlana [45]

Answer:

B = CHCl2 + Cl2 --> CHCl3 + Cl

Explanation:

Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.

Free radical chlorination is divided into 3 steps which are:

The initiation step

The propagation step

The termination step

So in reference to the question, propagation step involves two steps.

The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.

The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.

You would find in the attachment the 2 step mechanism.

3 0

2 years ago

The highest energy occupied molecular orbital in the f-f bond of the f2 molecule is _____

Morgarella [4.7K]

The basis of finding the answer to this problem is to know the electronic configuration of Fluorine. That would be: [He] 2s² 2p⁵. The valence electrons, which are the outermost electrons of the atom, are the ones that participate in bonding. Since the highest orbital for F is 2p, that means the highest energy occupied would be 2.

3 0

2 years ago

A gas sample occupies 3.50 liters of volume at 20.°c. what volume will this gas occupy at 100.°c (reported to three significant

VLD [36.1K]

Explanation:

According to Charle's law, at constant pressure the volume of an ideal gas is directly proportional to the temperature.

That is, $Volume \propto Temperature$

Hence, it is given that $V_{1}$ is 3.50 liters, $T_{1}$ is 20 degree celsius, and $T_{2}$ is 100 degree celsius.

Therefore, calculate $V_{2}$ as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

$\frac{3.50 liter}{20^{o}C} = \frac{V_{2}}{100^{o}C}$

$V_{2}$ = 17.5 liter

Thus, we can conclude that volume of gas required at 100 degree celsius is 17.5 liter.

6 0

2 years ago

Read 2 more answers

Diborane, B2H6 a possible rocket propellant, can be made by using lithium hydride (LiH): 6 LiH+ 2 BCl2àB2H6+ 6 LiCl . If you mix

Genrish500 [490]

Answer :

(a) Limiting reactant = $LiH$

(b) The excess reactant = $BCl_3$

(c) The percent of excess reactant is, 50.87 %

(d) The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

(e) The mass of $LiCl$ produced is, 1066.42 lb

Explanation : Given,

Mass of $LiH$ = 200 lb = 90718.5 g

conversion used : (1 lb = 453.592 g)

Mass of $BCl_3$ = 1000 lb = 453592 g

Molar mass of $LiH$ = 7.95 g/mole

Molar mass of $BCl_3$ = 117.17 g/mole

Molar mass of $B_2H_6$ = 27.66 g/mole

Molar mass of $LiCl$ = 42.39 g/mole

First we have to calculate the moles of $LiH$ and $BCl_3$ .

$\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=\frac{90718.5g}{7.95g/mole}=11411.13moles$

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{453592g}{117.17g/mole}=3871.23moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$6LiH+2BCl_3\rightarrow B_2H_6+6LiCl$

From the balanced reaction we conclude that

As, 6 moles of $LiH$ react with 1 mole of $BCl_3$

So, 11411.13 moles of $LiH$ react with $\frac{11411.13}{6}=1901.855$ moles of $BCl_3$

From this we conclude that, $BCl_3$ is an excess reagent because the given moles are greater than the required moles and $LiH$ is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 3871.23 - 1901.855 = 1969.375 moles

Total excess reactant = 3871.23 moles

Now we have to determine the percent of excess reactant $(BCl_3)$ .

$\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100$

$\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%$

The percent of excess reactant is, 50.87 %

Now we have to calculate the moles of $B_2H_6$ .

As, 6 moles of $LiH$ react to give 1 mole of $B_2H_6$

So, 11411.13 moles of $LiH$ react to give $\frac{11411.13}{6}=1901.855$ moles of $B_2H_6$

Now we have to calculate the mass of $B_2H_6$ .

$\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6$

$\text{Mass of }B_2H_6=(1901.855mole)\times (27.66g/mole)=52605.3093g$

Now we have to calculate the percent yield of $B_2H_6$ .

$\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%$

The percent yield of $B_2H_6$ or percent conversion of $LiH$ to $B_2H_6$ is, 38.80 %

Now we have to calculate the moles of $LiCl$ .

As, 6 moles of $LiH$ react to give 6 mole of $LiCl$

So, 11411.13 moles of $LiH$ react to give 11411.13 moles of $LiCl$

Now we have to calculate the mass of $LiCl$ .

$\text{Mass of }LiCl=\text{Moles of }LiCl\times \text{Molar mass of }LiCl$

$\text{Mass of }LiCl=(11411.13mole)\times (42.39g/mole)=483717.8007g=1066.42lb$

The mass of $LiCl$ produced is, 1066.42 lb

8 0

2 years ago