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Sati [7]
3 years ago
5

Name two common compounds that contain sodium

Chemistry
1 answer:
Leviafan [203]3 years ago
5 0
Carbonate and <span>sulfate contain sodium (I belive) </span>
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Be sure to answer all parts. For the complete redox reactions given here, write the half-reactions and identify the oxidizing an
OlgaM077 [116]

Answer : O_2 is the oxidizing agent and Fe is the reducing agent.

Explanation :

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The balanced redox reaction is :

4Fe+3O_2\rightarrow 2Fe_2O_3

The half oxidation-reduction reactions are:

Oxidation reaction : Fe\rightarrow Fe^{3+}+3e^-

Reduction reaction : O_2+4e^-\rightarrow 2O^{2-}

In order to balance the electrons, we multiply the oxidation reaction by 4 and reduction reaction by 3 then added both equation, we get the balanced redox reaction.

Oxidation reaction : 4Fe\rightarrow 4Fe^{3+}+12e^-

Reduction reaction : 3O_2+12e^-\rightarrow 6O^{2-}

4Fe+3O_2\rightarrow 2Fe_2O_3

In this reaction, 'Fe' is the reducing agent that loses an electron to another chemical species in a redox chemical reaction and itself gets oxidized and 'O_2' is the oxidizing agent that gain an electron to another chemical species in a redox chemical reaction and itself gets reduced.

Thus, O_2 is the oxidizing agent and Fe is the reducing agent.

7 0
3 years ago
How many helium atoms are there in a helium blimp containing 539kg of helium
MaRussiya [10]
Atomic mass of helium is 4.002642g/mol 
(542000g)/(4.002642g/mol)*6.02*10^23 = 8.15*10^28 atoms
3 0
3 years ago
Read 2 more answers
Plz help if you can
Vlad [161]

Answer: bro pic is not cleyer

7 0
3 years ago
A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 6-mL
scoray [572]

Answer:

(A) 4.616 * 10⁻⁶ M

(B) 0.576 mg CuSO₄·5H₂O

Explanation:

  • The molar weight of CuSO₄·5H₂O is:

63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol

  • The molarity of the first solution is:

(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M

The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.

  • Now we use the dilution factor in order to calculate the molarity in the second solution:

(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M

To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:

  • 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
  • 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
5 0
3 years ago
What does percent composition tell you about a molecule?
GaryK [48]

Answer:

C

Explanation:

3 0
2 years ago
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