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azamat
2 years ago
6

In a redox reaction, which ion would you expect to be more easily reduced? In a redox reaction, which ion would you expect to be

more easily reduced? Hypochlorite ion, ClO− They are equally reducible. Not enough information is given. Perchlorate, ClO−4
Chemistry
1 answer:
Levart [38]2 years ago
7 0

Answer:

In a redox reaction, hypochlorite ion would be more easily reduced.

Explanation:

Hypochlorite compounds are generally unstable - e.g. sodium hypochlorite is not available in solid form, because removing water from a NaClO disolution, causes a dismutation, turning it in a mixture of NaCl y NaClO3. Heating the same disolutions will also cause this effect. Hypochlorite will decompose under solar light into chlorides and oxygen.

Due to their low stability, hypochlorite compounds (i.e. those containing the hypochlorite ion) are very strong oxidants. Perchlorate ion can balance its negative charge instead and thus will be less prone to reduction than hypochlorite.

You might be interested in
The density of ethanol is 0.789 g/mL. Calculate the volume of 600 g of ethanol.
Anettt [7]

Answer:

The answer is

<h2>760.46 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}  \\

From the question

mass of ethanol = 600 g

density = 0.789 g/mL

It's volume is

volume =  \frac{600}{0.789}  \\  = 760.4562737...

We have the final answer as

<h3>760.46 mL</h3>

Hope this helps you

6 0
3 years ago
Which isotope is used to treat cancer?<br> (1) C-14 (3) Co-60<br> (2) U-238 (4) Pb-206
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I believe doctors use (3) Co-60 to treat ILWs. Uranium and Carbon are not used, and a different isotope of lead is used to be attached to monoclonal antibodies.
8 0
3 years ago
How many formula units are in 6.12 kg of lithium nitride
BartSMP [9]
There is 213 units i believe hope so
7 0
3 years ago
Several years ago, a group of scientists discovered evidence for a new element not found on the periodic table. Most
MatroZZZ [7]

<em>It is beneficial because it provides stronger support for the evidence related to the discovery</em>

<u>Answer:</u> <em>The correct answer is the second option that is given with shows it is beneficial.</em>

<u>Explanation:</u>

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3 0
2 years ago
85 POINTS PLEASE HELP ASAP
serious [3.7K]
1. Answer is: there are 1.41·10²³ molecules of oxygen.
1) calculate amount of substance for oxygen gas:
m(O₂) = 7.5 g; mass of oxygen.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 7.5 g ÷ 32 g/mol.
n(O₂) = 0.234 mol.
2) calculate number of molecules:
N(O₂) = n(O₂) · Na.
N(O₂) = 0.234 mol · 6.022·10²³ 1/mol.
N(O₂) = 1.41·10²³.
Na - Avogadro constant.

2. Answer is: the percent yield for the reaction is 61.77%.
Balanced chemical reaction: 2HgO(s) → 2Hg(l) + O₂<span>(g).
m(HgO) = 4.37 g.
n</span>(HgO) = n(HgO) ÷ M(HgO).
n(HgO) = 4.37 g ÷ 216.6 g/mol.
n(HgO) = 0.02 mol.
From chemical reaction: n(HgO) : n(Hg) = 2 : 2 (1 :1).
n(Hg) = n(HgO) = <span>0.02 mol; amount of substance.
m</span>(Hg) = n(Hg) ·M(Hg).
m(Hg) = 0.02 mol · 200.6 g/mol.
m(Hg) = 4.047 g.
yield = 2.5 g ÷ 4.047 g · 100%.
<span>yield = 61.77%.

3. Answer is: 68.16  </span><span>grams of the excess reactant (oxygen) remain.
</span>Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃<span>(g).
m(Fe) = 27.3 g.
n</span>(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 79.9 g.
n(O₂) = 79.9 g ÷ 32 g/mol.
n(O₂) = 2.497 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
0.489 mol : n(O₂) = 4 : 3.
n(O₂) = 3 · 0.489 mol ÷ 4.
n(O₂) = 0.367 mol.
Δn(O₂) = 2.497 mol - 0.367 mol.
Δn(O₂) = 2.13 mol.
m(O₂) = 2.13 mol · 32 g/mol.
m(O₂) = 68.16 g.

4. Answer is: there are 0.603 moles of ammonia.
m(NH₃) = 10.25 g; mass of ammonia.
M(NH₃) = Ar(N) + 3Ar(H) · g/mol.
M(NH₃) = 14 + 3·1 · g/mol.
M(NH₃) = 17 g/mol; molar mass of ammonia.
n(NH₃) = m(NH₃) ÷ M(NH₃).
n(NH₃) = 10.25 g ÷ 17 g/mol.
n(NH₃) = 0.603 mol; amount of substance (ammonia).

5. Answer is: the empirical formula mass of P₂O₅ is 141.89.
<span>Empirical formula gives the proportions of the elements present in a compound.
</span>Atomic mass of phosphorus is 30.97 g/mol.
Atomic mass of oxygen is 15.99 g/mol.
In phosphorus (V) oxide there are two atoms of phosphorus and five atoms of oxygen:
EFM(P₂O₅) = 2·30.97 g/mol + 5·15.99 g/mol = 141.89 g/mol. 

6. Answer is: there are 1.108·10²⁴ molecules of water.
n(H₂O) = 1.84 mol; amount of substance (water).
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 1.84 mol · 6.022·10²³ 1/mol.
N(H₂O) = 11.08·10²³.
N(H₂O) = 1.108·10²⁴.
Na - Avogadro constant (<span>number of particles (ions,</span> atoms<span> or </span>molecules), that are contained in <span>one </span>mole of substance<span>).
</span>
7. Answer is: iron (Fe) <span>is the limiting reactant.
</span>Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 45.8 g.
n(O₂) = 45.8 g ÷ 32 g/mol.
n(O₂) = 1.431 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
For 1.431moles of oxygen we need:
1.431 mol : n(Fe) = 3 : 4.
n(Fe) = 1.908 mol, there is no enough iron.

8. Answer is: there are 0.435 moles of C₆H₁₄.<span>
N(C₆H₁₄) = 2.62·10²³; number of molecules.</span><span>
n(C₆H₁₄) = N(C₆H₁₄) ÷ Na.</span><span>
n(C₆H₁₄) = 2.62·10²³ ÷ 6.022·10²³ 1/mol.</span><span>
n(C₆H₁₄) = 0.435 mol; amount of substance of </span>C₆H₁₄.<span>
Na - Avogadro constant or Avogadro number.
</span>
9. Answer is: 3.675 <span>moles of carbon(II) oxide are required to completely react.
</span>Balanced chemical reaction: Fe₂O₃<span>(s) + 3CO(g) ⟶ 2Fe(s) + 3CO</span>₂<span>(g).
n(</span>Fe₂O₃) = 1.225 mol; amount of substance.
From chemical reaction: n(Fe₂O₃) : n(CO) = 1 : 3.
1.225 mol : n(CO) = 1 : 3.
n(CO) = 3 · 1.225 mol.
n(CO) = 3.675 mol.

10. Answer is: there are 2.158 moles of barium atoms.<span>
N(Ba</span><span>) = 2.62·10²³; number of atoms of barium.
n</span>(Ba) = N(Ba)<span> ÷ Na.
n</span>(Ba) = 1.3·10²⁴<span> ÷ 6.022·10²³ 1/mol.
n</span>(Ba)<span> = 2.158 mol; amount of substance of barium</span>.<span>
Na - Avogadro constant or Avogadro number.

11. Answer is: </span>6.26·10²³ <span>carbon atoms are present.
</span>n(C₂H₆O) = 0.52 mol; amount of substance.<span>
N</span>(C₂H₆O) = n(C₂H₆O) · Na.<span>
N</span>(C₂H₆O) = 0.52 mol · 6.022·10²³ 1/mol.<span>
N</span>(C₂H₆O) = 3.13·10²³.<span>
In one molecule of </span>C₂H₆O there are two atoms of carbon:<span>
N(C</span>) = N(C₂H₆O) · 2.
N(C) = 3.13·10²³ · 2.
N(C) = 6.26·10²³.
<span>
12. Answer is: </span><span>the empirical formula is C</span>₂H₄O.<span>
</span><span>If we use 100 grams of compound:
</span>1) ω(C) = 51% ÷ 100% = 0.51.
m(C) = ω(C) · m(compound).
m(C) = 0.51 · 100 g.
m(C) = 51 g.
n(C) = m(C) ÷ M(C).
n(C) = 51 g ÷ 12 g/mol.
n(C) = 4.25 mol.
2) ω(H) = 9.3 % ÷ 100% = 0.093.
m(H) = 0.093 · 100 g.
m(H) = 9.3 g.
n(H) = 9.3 g ÷ 1 g/mol.
n(H) = 9.3 mol
3) ω(O) = 39.2 % ÷ 100%.
ω(O) = 0.392.
m(O) = 0.392 · 100 g.
m(O) = 39.2 g.
n(O) = 39.2 g ÷ 16 g/mol.
n(O) = 2.45 mol.
4) n(C) : n(H) : n(O) = 4.25 mol : 9.3 mol : 2.45 mol /2.45 mol.
n(C) : n(H) : n(O) = 1.73 : 3.795 : 1.
8 0
2 years ago
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