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3 years ago

Iron (III) oxide pentahydrate formula

Chemistry

1 answer:

Nina [5.8K]3 years ago

4 0

Answer:

Molecular Formula Fe2O12S3·5H2O

IUPAC Name iron(3+);tri sulfate;pentahydrate

Explanation:

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1. Calculate the energy of a photon with a frequency of 3x1015 Hz. ?????

Afina-wow [57]

Answer:

1.99 x 10⁻¹⁸J

Explanation:

Given parameters:

Frequency of the wave = 3 x 10¹⁵Hz

Unknown:

Energy of the photon = ?

Solution:

To solve this problem, we use the expression below;

E = hf

Where E is the energy, h is the Planck's constant and f is the frequency

Now insert the parameters and solve for E;

E = 6.63 x 10⁻³⁴ x 3 x 10¹⁵ = 19.9 x 10⁻¹⁹J or 1.99 x 10⁻¹⁸J

6 0

3 years ago

What volume of a 6.67 M NaCl solution contains 3.12 mol NaCl? L

harkovskaia [24]

Answer:

0.47dm³

Explanation

Given parameters :

Molarity of NaCl = 6.67M

Number of moles = 3.12mol

Volume of NaCl =?

Volume of NaCl = number of moles/Molarity

Volume of NaCl = 3.12mol/6.67M

Volume of NaCl = 0.47dm³

4 0

3 years ago

Read 2 more answers

Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit

ser-zykov [4K]

Answer:

For a: The edge length of the unit cell is 314 pm

For b: The radius of the molybdenum atom is 135.9 pm

Explanation:

For a:

To calculate the edge length for given density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density = $10.28g/cm^3$

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell =?

Putting values in above equation, we get:

$10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm$

Conversion factor used: $1cm=10^{10}pm$

Hence, the edge length of the unit cell is 314 pm

For b:

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

$R=\frac{\sqrt{3}a}{4}$

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

$R=\frac{\sqrt{3}\times 314}{4}=135.9pm$

Hence, the radius of the molybdenum atom is 135.9 pm

4 0

3 years ago

Is this an example of decomposition 2CH4 + 402 → 2C02 + 4H2O

Molodets [167]

Answer:

Identify each equation as a composition reaction, a decomposition reaction, or neither.

Fe2O3 + 3 SO3 → Fe2(SO4)3

NaCl + AgNO3 → AgCl + NaNO3

(NH4)2Cr2O7 → Cr2O3 + 4 H2O + N2

Solution

In this equation, two substances combine to make a single substance. This is a composition reaction.

Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.

A single substance reacts to make multiple substances. This is a decomposition reaction.

Test Yourself

Identify the equation as a composition reaction, a decomposition reaction, or neither.

C3H8 → C3H4 + 2 H2

Explanation:

I hope I help :)))

5 0

3 years ago

Why do different materials have similar properties

Deffense [45]

Answer:

The difference in the number of protons and neutrons in atoms account for many of the different properties of elements.

7 0

2 years ago

Read 2 more answers