Answer:
decelerating hope this helps❤️
When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.
NO2→NO^−3
NO2→NO
and do the usual changes
First, balance the two half reactions:
3. NO2 +H2O →NO^−3 + 2 H^+ + e−
4. NO2 +2 H^+ + 2e− → NO + H2O
Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:
5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−
Now add Eqn 4 and 5 (the electrons now cancel each other):
3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+
and cancel terms that’s common to both sides:
3NO2 + H2O → NO + 2NO^−3 + 2H+
This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.
Learn more about balancing equation here:
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<span> When an </span>acid and a base<span> are placed together, they </span>react<span> to neutralize the </span>acid<span> and </span>base<span> properties, producing a salt. The H(+) cation of the </span>acid<span>combines with the OH(-) anion of the </span>base<span> to form water.</span>
Answer:
Here's what I get.
Explanation:
At the end of the reaction you will have a solution of the alcohol in THF.
The microdistillation procedure will vary, depending on the specific apparatus you are using, but here is a typical procedure.
- Transfer the solution to a conical vial.
- Add a boiling stone.
- Attach a Hickman head (shown below) and condenser.
- Place the assembly in in the appropriate hole of an aluminium block on top of a hotplate stirrer.
- Begin stirring and heating at a low level so the THF (bp 63 °C) can distill slowly.
- Use a Pasteur pipet to withdraw the THF as needed.
- When all the THF has been removed, raise the temperature of the Al block and distill the alcohol (bp 143 °C).
Answer:
8.934 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 192.12 44.01
H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂
m/g: 13.00
For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.
(a) Calculate the <em>moles of H₃Cit
</em>
n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)
n = 0.067 67 mol H₃Cit
(b) Calculate the <em>moles of CO₂
</em>
The molar ratio is (3 mol CO₂/1 mol H₃Cit)
n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)
n = 0.2030 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
m = 8.934 g CO₂
