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Anna35 [415]
3 years ago
9

Write empirical formula

Chemistry
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

Fe(MnO_{4})_{2}\\\\ Fe(IO_{3})_{2}\\\\NH_{4}MnO_{4}\\\\NH_{4}IO_{3}

Explanation:

Fe^{2+}(MnO_{4}^{-})_{2}--->Fe(MnO_{4})_{2}\\\\Fe^{2+}(IO_{3}^{-})_{2}---> Fe(IO_{3})_{2}\\\\NH_{4}^{+}MnO_{4}^{-}--->NH_{4}MnO_{4}\\\\NH_{4}^{+}IO_{3}^{-}--->NH_{4}IO_{3}

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Help, cant get it. Answers in picture
Aleks04 [339]

Answer:17.0

Explanation:

8 0
3 years ago
A sample of hydrogen gas at a pressure of 0.926 atm and a temperature of 29.5 C, occupies a volume of 457 mL. If the gas is allo
klio [65]

Answer:

V₂ = 946.72 mL

Explanation:

Given data;

Initial pressure = 0.926 atm

Initial volume = 457 mL

Temperature = constant = 29.5°C

Final pressure = 0.447 atm

Final volume = ?

Solution:

The given problem will be solved through the Boyle's law,

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

by putting values,

P₁V₁ = P₂V₂

0.926 atm × 457 mL = 0.447 atm × V₂

V₂ = 423.18 atm. mL/ 0.447 atm

V₂ = 946.72 mL

4 0
3 years ago
The radius of a helium atom is a 31 PM. What is the radius in nanometers?
Paladinen [302]

Answer:

Thus, the radius of the helium atom in nanometers is - 0.031 nm

Explanation:

Given that:-

The radius of the helium atom = 31 pm

Considering the conversion of length in pm to the length in nm as:-

1 pm = 0.001 nm

So,

Applying the above conversion factor in the radius of helium atom as:-

Radius = 31\times 0.001 nm = 0.031 nm

<u>Thus, the radius of the helium atom in nanometers is - 0.031 nm</u>

7 0
3 years ago
How is radiation different from conduction?
steposvetlana [31]

Answer:

<u>Radiation is the transfer of energy by waves, and conduction is the transfer of heat through contact with air.</u>

Explanation:

Conduction is the transfer of thermal energy through direct contact.  Radiation is the transfer of thermal energy through thermal emission.

7 0
3 years ago
How do you work out question 1a?
Sliva [168]

Answer:

-125 kJ

Explanation:

You calculate the energy required to break all the bonds in the reactants. Then you subtract the energy to break all the bonds in the products.

                     H₂C=CH₂   +    H₂ ⟶    H₃C-CH₃

Bonds:       4C-H + 1C=C     1H-H     6C-H + 1C-C

D/kJ·mol⁻¹:  413       612        436       413      347

The formula relating ΔHrxn and bond dissociation energies (D) is

ΔHrxn = Σ(Dreactants) – Σ(Dproducts)

(Note: This is an exception to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)

<em>For the reactant</em>s:

Σ(Dreactants) = 4 × 413 + 1 × 612 + 1 × 436 = 2700 kJ

<em>For the products:</em>

Σ(Dproducts) = 6 × 413 + 1 × 347 = 2825 kJ

<em>For the system</em> :

ΔHrxn = 2700 - 2825 = -125 kJ

4 0
3 years ago
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