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yan [13]
3 years ago
14

10. How many moles of Fe2+ can be oxidized to Fe+ by 0.75 mole of Cl2, according to

Chemistry
1 answer:
Nat2105 [25]3 years ago
4 0
It means 3 moles of Fe2+ are being oxidized to Fe3+.

3Fe2+ → Fe3+ ; so balance that out you get 3Fe2+ → 3Fe3+ + 3e- (1)

Therefore,from the balanced equation you only need 1 mole of Cl2 to oxidize 3moles of Fe2+ .
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3 years ago
Substances that are likely to disassociate in water
mr_godi [17]

Answer:

sugar,dirt,spit

Explanation:

5 0
3 years ago
Describe a method to calculate the average atomic mass of the sample in the previous question using only the atomic masses of li
alexira [117]

Answer:

Explanation:

To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.

The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.

This can be expressed below:

        RAM = Σmₙαₙ

where mₙ is the mass of isotope n

           αₙ is the abundance of isotope n

for this problem:

RAM of Li = m₆α₆ + m₇α₇

       m₆ is mass of isotope Li-6

        α₆ is the abundance of isotope Li-6

       m₇ is mass of isotope Li-7

        α₇ is the abundance of isotope Li-7

3 0
3 years ago
What is the temperature of a gas that is expanded from 3.75 L at 37 degrees Celsius to 5.6 L?
deff fn [24]

Answer:

190 °C  

Step-by-step explanation:

The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:  

V₁/T₁ = V₂/T₂      Invert both sides of the equation.  

T₁/V₁ = T₂/V₂      Multiply each side by V₂

T₂ = T₁ × V₂/V₁

=====

V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K  

V₂ = 5.6 L;   T₂ = ?  

=====

T₂ = 310.15 × 5.6/3.75

T₂ = 310.15 × 1.49

T₂ = 463 K

t₂ = 463 – 273.15

t₂ = 190 °C

3 0
3 years ago
The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?
Agata [3.3K]

<u>Answer:</u> The molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

PbI_2\rightleftharpoons Pb^{2+}+2I^-

                s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Pb^{2+}][I^-]^2

We are given:

K_{sp}=7.9\times 10^{-9}

Putting values in above equation, we get:

7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L

Hence, the molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

3 0
3 years ago
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