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Travka [436]
3 years ago
11

9 How many grams of O₂ are needed to produce 15.5 g Fe₂O3 in the following reaction? Fe(s) + O₂(g) → Fe₂O3 (s)​

Chemistry
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer:

Explanation:

so u can work out the amount of moles in FeO3 by doing mr of fe3o3 is 55.8*3+16*3=215.4

moles= mass/mr so you do 15.5g/215.4=0.0719 moles

then using 1 to 1 ratio so O2 moles is 0.0719

then use the equation mass=mole*mr

so 0.0719*16=1.15g

hope this make sense :)

You might be interested in
How many moles of CoH12O6 are consumed<br> if 6 moles of O2 are consumed?
Rashid [163]

Answer:

2

Explanation:

There are 3 moles O2 in 1 mole CoH12O6 so 2 moles are consumed

7 0
3 years ago
When an electron passes through the magnetic field of a horseshoe magnet, the electron's?
blsea [12.9K]

When an electron passes through the magnetic field of a horseshoe magnet, the electron's direction is changed.

Path of an electron in a magnetic field

The force (F) on wire of length L carrying a current I in a magnetic field of strength B is given by the equation:

F = BIL

But Q = It and since Q = e for an electron and v = L/t you can show that :

Magnetic force on an electron = BIL = B[e/t][vt] = Bev where v is the electron velocity

In a magnetic field the force is always at right angles to the motion of the electron (Fleming's left hand rule) and so the resulting path of the electron is circular.

Therefore :

Magnetic force = Bev = mv2/r = centripetal force

v = [Ber]/m

and so you can see from these equations that as the electron slows down the radius of its orbit decreases.

If the electron enters the field at an angle to the field direction the resulting path of the electron (or indeed any charged particle) will be helical. Such motion occurs above the poles of the Earth where charges particles from the Sun spiral through the Earth's field to produce the aurorae.

To learn more about electron : brainly.com/question/860094

#SPJ4

4 0
2 years ago
Remical Reactions 8.P.13 1 8 of 40
ruslelena [56]

Answer:

H₂O + CO₂      →    H₂CO₃

Option D is correct.

Law of conservation of mass:

According to this law, mass can neither be created nor destroyed in a chemical equation.

This law was given by French chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Now we will apply this law to given chemical equations:

A) H₂ + O₂   →    H₂O

There are two H and two O atoms present on left side while on right side only one O and two H atoms are present so mass in not conserved. This option is incorrect.

B) Mg + HCl   →   H₂ + MgCl₂

In this equation one Mg, one H and one Cl atoms are present on left side of equation while on right side two H, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.

C) KClO₃      →     KCl + O₂

There are one K, one Cl and three O atoms are present on left side of equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.

D)  H₂O + CO₂      →    H₂CO₃

There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. This option is correct.

5 0
3 years ago
Can someone please help me with the correct formula for this?
Anarel [89]

Answer:

C₂H₄O

CH₃CHO

Explanation:

I'm not sure if you want the molecular formula or the condensed structure, but I will give you both.

Molecular formula:

You have 2 carbons (C₂), 4 hydrogens (H₄), and 1 oxygen (O).  The molecular formula will be C₂H₄O.

Condensed Structure:

You have a carbon bonded to three hydrogens (CH₃).  This carbon is bonded to a carbon that is bonded to a hydrogen and oxygen (CHO).  The condensed structure will be CH₃CHO.

6 0
3 years ago
The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

A=6R^{2}\sqrt{3}

Here, R is radius and is related to a as follows:

R=\frac{a}{2}

Putting the value in expression for area,

A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

The value of a is 0.4961 nm

Since, 1 nm=10^{-7}cm

Thus, 0.4961 nm=4.961\times 10^{-8} cm

Putting the value,

Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}

Now, volume can be calculated as follows:

V=Area\times c

The value of c is 1.360 nm or 1.360\times 10^{-7} cm

Putting the value,

V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}

now, number of atom in unit cell can be calculated by using the following formula:

n=\frac{\rho N_{A}V_{c}}{A}

Here, A is atomic mass of Cr_{2}O_{3} is 151.99 g/mol.

Putting all the values,

n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18

Thus, there will be 18 Cr_{2}O_{3} units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of Cr^{3+} and O^{2-} is 62 pm and 140 pm respectively.

Converting them into cm:

1 pm=10^{-10}cm

Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

Putting the values,

V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758

Thus, atomic packing factor is 0.758.

6 0
3 years ago
Read 2 more answers
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