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Liono4ka [1.6K]
3 years ago
7

You used a calorimeter in the heat transfer lab. Explain how the calorimeter works, and how to calculate the heat given off or a

bsorbed by the reaction being studied.
Chemistry
2 answers:
Olenka [21]3 years ago
7 0
A calorimeter contains reactants and a substance to absorb the heat absorbed. The initial temperature (before the reaction) of the heat absorbent is measured and then the final temperature (after the reaction) is also measured. The absorbent's specific heat capacity and mass are also known. Given all of this data, the equation:
Q = mcΔT 
To find the heat released.
notka56 [123]3 years ago
3 0

Calorimeter functions by possessing a known mass of familiar substance combust or react in an enclosed space. The calorimeter exhibits an agent for captivation of the heat discharged at the time of reaction or combustion, for example, water may act as the heat absorbing agent.  

The variation in temperature of the heat absorbent in the company of its mass and specific heat capacity are utilized to find out the energy discharged with the help of the equation:  

Q = mCΔT

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There are ________ σ bonds and ________ π bonds in h3c-ch2-ch=ch-ch2-c≡ch.
FrozenT [24]
The number of sigma and pi bonds are,

          Sigma Bonds  =  16

          Pi Bonds         =   3

Explanation:
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Below all black bonds are sigma bonds, while in alkene there is one pi bond and in alkyne there are two pi bonds.

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3 years ago
How many grams of water are produced if 3.2 moles of hydrogen reacts
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Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

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

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Vadim26 [7]
The precipitation reaction of sodium nitrate and lead (II) chloride would not happen. No reaction will happen between these substances. Hope this answers the question. Have a nice day.
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