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ch4aika [34]
3 years ago
8

For the chemical reaction hclo4(aq)+naoh(aq)⟶h2o(l)+naclo4(aq) write the net ionic equation, including the phases.

Chemistry
1 answer:
mojhsa [17]3 years ago
5 0
  The net   ionic equation  is as below

 H^+(aq)  +  H^-(aq)  →  H2O(l)  

         Explanation
from the chemical equation

HClO4(aq)  + NaOH (aq)  →    H2O(l)  + NaClO4(aq)

write  ionic equation
that is, 
H^+(aq)+ClO4^-(aq) + Na^+(aq) +OH^-(aq)→ H2O(l) + Na^+(aq)+ClO4^-(aq)

cancel the spectator  ions   in both side(ions that do no take place in  a chemical equation)
that  ClO4^-   and Na^+

the net  ionic is therefore
= H^+(aq)  +  OH^-(aq)   →  H2O(l)

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Lead(II) oxide from an ore can be reduced to elemental lead by heating in a furnace with carbon. Calculate the expected yield of
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Answer:

53kg is the expected yield of lead

Explanation:

Firstly, in order to solve this question, we need to write the equation of reaction correctly. This is as follows:

PbO(s) + C(s) ---> Pb(l) + CO(g)

We proceed from here. We should get the limiting reactant but this can only be obtained by getting the number of moles of each reactant present.

The formula to use across all boards is that the number of moles is the mass of each of the reactant divided by the molar mass of each of the reactant.

For PBO, mass is 57kg = 57000g

Molar mass of PBO = 223.20g/mol

The number of moles is thus 57,000/223.2 = 255.37 moles

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Number of moles of carbon = 57000/12 = 4750 moles

From the number of moles, we can see that the number of moles of Carbon is greater than that of PbO. This means that PbO is the limiting reagent.

Hence we use it to calculate percentage yield.

The number of moles of lead formed is the same of number of moles of lead oxide = 255.37 since we have mole ratio of 1 to 1

The molar mass of lead is 207.20g/mol

The mass of lead formed is = moles of lead formed * molar mass of lead = 207.20 * 255.37 = 52,912g which is approximately 53kg

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6 0
3 years ago
If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?
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