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Harrizon [31]
3 years ago
9

Find the oxidizing agent and the reducing agent.

Chemistry
1 answer:
Vladimir [108]3 years ago
8 0
Oxidizing agent is that which is reduced and the reducing agent is that which is oxidized. Reduced is when the charged is decreased and oxidized when the charge is increased.

(1)  2Na + 2H2O(l) --> 2NaOH(aq) + H2(g)
The charge of Na in the reactant is 0 and the charge of Na in the NaOH is +1. Na is oxidized. Hence, it is the reducing agent. 

The charge of H in H2O is +1 while that in H2 is 0. H is reduced. Hence, it is the oxidizing agent.

(2)  C(s) + O2(g) --> CO2(g)

The charge of C in the reactant side is 0 and that its charge in CO2 is +4. C is oxidized. Hence, it is the reducing agent. 

The charge of O in O2 is 0 while in CO2, its charge is -2. O is reduced. Hence, it is the oxidizing agent.

(3)  2MnO⁻⁴ + SO2 + 2H2O --> 2Mn²⁺ + 5SO2⁻⁴ 4H⁺

The charge of Mn in MnO⁻⁴ is 4+ while its charge in Mn²⁺ is 2+. Mn is reduced. Hence, it is the oxidizing agent.

The charged of S in SO2 is -4 while its charge in SO₂⁻⁴ is 0. S is oxidized. Hence, it is the reducing agent. 
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Answer:

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Explanation:

The 1-ethyl is because you start numbering from the longest branch, towards the next closest branch. Prefix "eth-" means two, there are 2 carbons in the longest branch. 3-methyl is because the next branch is at number 3, and prefix "meth-" means 1, there is 1 carbon in that chain. "Cyclo" in cyclohexane means the skeletal model is shaped like a ring, and the "hexane" means there are 6 carbons in the ring. Prefix "hex" means 6.

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What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?
Ilia_Sergeevich [38]

Answer: The volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of O_{2} gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of O_{2} (molar mass = 32.0 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L

Thus, we can conclude that the volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

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