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Advocard [28]
2 years ago
12

How many moles of iron are there in 55.85g of Fe3O4

Chemistry
1 answer:
soldier1979 [14.2K]2 years ago
8 0

Answer:

• Molecular mass of Iron (III) tetraoxide

\dashrightarrow \: { \tt{(56 \times 3) + (16 \times 4)}} \\  = { \tt{168 + 64}} \\  = { \tt{232\:g}}

[ molar masses: Fe → 56, O → 16 ]

\dashrightarrow \:{ \rm{232 \: g \:  = 1 \: mole}} \\ \\   \dashrightarrow \: { \rm{55.85 \: g = ( \frac{55.85}{232}) \: moles }} \\  \\ \dashrightarrow \:  { \boxed{ \tt{ = 0.24 \: moles}}}

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nignag [31]

Organic chemistry as the the study of general properties and compositions of organic compounds.

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<h3>What is organic chemistry?</h3>

Organic chemistry can be simply defined as the study of organic compounds.

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6 0
2 years ago
Read 2 more answers
Column A. (4pts/.5 pts each) Order the layers from youngest to oldest. Write the letter for each layer in the correct order in t
vladimir1956 [14]

Answer:

principle of superposition

8 0
2 years ago
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

PCl_3,p_1'=6.798 Torr

Cl_2,p_2'=26.398 Torr

PCl_5,p_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=p_1=13.2 Torr

Partial pressure of the Cl_2=p_2=13.2 Torr

Partial pressure of the PCl_5=p_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{p_1}{p_1\times p_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=p_1'=13.2 Torr

Partial pressure of the Cl_2=p_2'=?

Partial pressure of the PCl_5=p_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=p_1'+p_2'+p_3'

263.0Torr=13.2 Torr+p_2'+217.0 Torr

p_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{p_3'}{p_1'\times p_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

6 0
2 years ago
What is the difference between an Isotope and an Allotrope?
Kaylis [27]
Allotropes are elements on the periodic table that have more than one crystalline form . There are three forms of the element carbon : Diamond, Graphite, and Fullerenes. Isotopes are atoms of the same element with the same atomic number but have a different mass number.
8 0
3 years ago
What is the unit of measurement for mass?
Licemer1 [7]
Grams. It is a smaller unit. 
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3 years ago
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