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zmey [24]
3 years ago
10

Aluminum metal reacts with aqueous cobalt(II) nitrate to form aqueous aluminum nitrate and cobalt metal. What is the stoichiomet

ric coefficient for cobalt when the chemical equation is balanced using the lowest whole-number stoichiometric coefficients?
Chemistry
1 answer:
Lelu [443]3 years ago
7 0

Answer:

the stoichiometric coefficient for cobalt is 3

Explanation:

the unbalanced reaction would be

Co(NO₃)₂+ Al → Al(NO₃)₃ + Co

One way to solve is to build a system of linear equations for each element (or group as NO₃) , knowing that the number of atoms of each element is conserved.

For smaller reactions a quick way to solve it can be:

- First the Co as product and as reactant needs to have the same stoichiometric coefficient

- Then the Al as product and as reactant needs to have the same stoichiometric coefficient

- After that we look at the nitrates . There are 2 as reactants and 3 as products . Since the common multiple is 6 then multiply the reactant by 3 and the product by 2.

Finally the balanced equation will be

3 Co(NO₃)₂+ 2 Al → 2 Al(NO₃)₃ + 3 Co

then the stoichiometric coefficient for cobalt is 3

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They use information from other things that relate to that topic and then use that to do their research
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Lithium diisopropylamide (LDA) is used as a strong base in organic synthesis. LDA is itself prepared by an acid-base reaction be
kakasveta [241]

Explanation:

Lithium diisopropylamide (LDA) is used in many organic synthesis and is a strong base. It is prepared by the acid base reaction of N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) and butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ).

The equation is show below as:

[(CH₃)₂CH]₂NH + Li⁺⁻CH₂CH₂CH₂CH₃  ⇒  [(CH₃)₂CH]₂N⁻Li⁺  + CH₃CH₂CH₂CH₃

N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) is a weaker acid and hence,  LDA ( [(CH₃)₂CH]₂N⁻Li⁺ ) is stronger base. (Weaker acid has stronger conjugate base)

Butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ) is a very strong base and hence, butane ( CH₃CH₂CH₂CH₃ ) is a very weak acid. (Strong base has weaker conjugate acid)

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3 years ago
Which model shows 6 electrons in the outer shell of the atom?
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D. Model number 2

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3 years ago
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What is the enthalpy of combustion (per mole) of C4H10 (g)? 
Artemon [7]
The balanced chemical reaction for the complete combustion of C4H10 is shown below:

                    C4H10 + (3/2)O2 --> 4CO2 + 5H2O

The enthalpy of formation are listed below:
          C4H10: -2876.9 kJ/mol
              O2:   none (because it is pure substance)
             CO2: -393.5 kJ/mol
             H2O: -285.8 kJ/mol

The enthalpy of combustion is computed by subtracting the total enthalpy formation of the reactants from that of the products.

               ΔHc = (4)(-393.5 kJ/mol) + (5)(-285.8 kJ/mol) - (-2876.9 kJ/mol)
                       = -<em>126.1 kJ</em>

Thus, the enthalpy of combustion of the carbon is -126.1 kJ. 
5 0
3 years ago
Read 2 more answers
Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation:
tekilochka [14]

Answer:

See explanation

Explanation:

First, let's write the balanced equation again:

2 NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)

Now, we know that the total pressure was 7.76 atm. This total pressure, is the sum of the pressure of water and CO2 like this:

Ptotal = Pwat + PCO2 (1)

This is the dalton's law for partial pressures.

The pressure can be also be relationed with the moles

Ratio of mole = Ratio of pressure

so, taking this in consideration we can say the following:

Pwater/PCO2 = moles water / moles CO2

As the only components exerting pressure are CO2 and Water (Because they are in gas phase), the total pressure can be splitted between the two of them so:

Pwater = Ptotal/2

Pwater = 7.76 / 2 = 3.88 atm

With this pressure, and using the ideal gas equation, we can know the moles of water:

PV = nRT

n = PV/RT     using R = 0.082 L atm / K mol

n = 3.88 * 5 / 0.082 * (160+273)

n = 0.546 moles of water

b) now that we have the moles of water, we can actually know the moles that reacted originally from the sodium carbonate by stechiometry.

2NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)    MMCO2 = 84 g/mol

the moles of NaHCO3 initially:

n = 100 / 84

n = 1.19 moles

so, If 1.19 moles of NaHCO3 reacted, and only produces 0.546 moles of water and CO2, then, the remaining moles of NaHCO3 is:

remaining moles = 1.19 - 0.546 = 0.644 moles

therefore the mass remaining:

mCO2 = 0.644 * 84

mCO2 = 54.096 g

c) As it was stated before, only the gaseous components are involved in the pressure, thus, in the kp expression which is:

Kp = Pwater * PCO2

Kp = 3.88 * 3.88

Kp = 15.0544

d) As the total pressure is 7.76 atm and the fact that NaHCO3 is solid, this component is not exerting any pressure in the reaction, as seen in the Kp expression, so it won't matter that if we raise a little the quantity of the reactant, it still has some remaining.

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3 years ago
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