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Alchen [17]
3 years ago
7

Upon combustion, a compound containing only carbon and hydrogen produces 2.28 gCO2 and 0.702 gH2O. Find the empirical formula of

the compound.
Chemistry
1 answer:
Alik [6]3 years ago
7 0

Answer:

EF = CH₂

Explanation:

The empirical formula of any compound would be the most simplified and simple chemical formula of a compound. To get this number, we need to get the data of how many moles of each atom of the compoud we have.

Analyzing the data, we just know that an unknown compound undergoes a combustion reaction and produces CO₂ and H₂O. As the compound only have Carbon and hydrogen, it's pretty easy to get the moles of each atom.

To get the moles of CO₂, we need the molar mass of CO₂, which is 44 g/mol, then the moles:

nCO₂ = 2.28 / 44 = 0.0518 moles of CO₂.

Now that we have the moles of CO₂, we can get the moles of Carbon. Since 1 mole of CO₂ is made of 1 mole of Carbon and 2 moles of oxygen, this means that we have 0.0518 moles of C.

Applying the same thing with the water, we have:

nH₂O = 0.702 / 18 = 0.039 moles of water

And 1 mole of water is made of 1 mole of oxygen and two moles of hydrogens, then the moles of Hydrogen are 0.078 moles.

With these moles, we can now calculate the empirical formula.

C = 0.0518 / 0.0518 = 1

H = 0.078 / 0.0518 = 1.51 ≅ 2

Now these two numbers, give us the number of atoms, in the empirical formula so:

<h2>EF = CH₂</h2>

Of course there is no compound that have only those number of atom, but remember that this is the empirical formula, so, it's a simplified or reduced formula of the molecular formula, which means that the real or whole molecular formula could have more atoms of C and H, but to know this, we need the molar weight of this compound, and the problem does not give this data.

Hope this helps

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andrew-mc [135]

Answer:

1280J are required.

Explanation:

Heat of fusion is defined as the amount of heat required to change its state from liquid to solid at its melting point at constant pressure.

As heat of fusion of gold is 64J/g, there are required 64J to melt 1g of gold at its melting point. The energy required to melt 20g is:

20g * (64J/g) =

1280J are required

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Use the given data at 500 K to calculate ΔG°for the reaction
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Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
3 years ago
One of relatively few reactions that takes place directly between two solids at room temperature is <img src="https://tex.z-dn.n
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Answer:

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b) 3.14g must be added

Explanation:

a) For the reaction:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)

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Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)

To balance hydrogens, the other coefficients are:

Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)

b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:

6.5 g × (1mol / 315.48g) =<em> 0.0206moles of  Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:

0.0206moles of  Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:

0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>

8 0
4 years ago
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ANEK [815]

Answer:

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