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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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6 0
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If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L
Tom [10]

Answer:

Approximately 2.53\; \rm L (rounded to three significant figures) assuming that {\rm HCl}\, (aq) is in excess.

Explanation:

When {\rm HCl} \, (aq) and {\rm AgNO_3}\, (aq) precipitate, {\rm AgCl} \, (s) (the said precipitate) and \rm HNO_3\, (aq) are produced:

{\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq) (verify that this equation is indeed balanced.)

Look up the relative atomic mass of \rm Ag and \rm Cl on a modern periodic table:

  • \rm Ag: 107.868.
  • \rm Cl: 35.45.

Calculate the formula mass of the precipitate, \rm AgCl:

\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm  g\cdot mol^{-1}\end{aligned}.

Calculate the number of moles of \rm AgCl formula units in 45.00\; \rm g of this compound:

\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}.

Notice that in the balanced equation for this reaction, the coefficients of {\rm AgNO_3} \, (aq) and {\rm AgCl}\, (s) are both one.

In other words, if {\rm HCl}\, (aq) (the other reactant) is in excess, it would take exactly 1\; \rm mol of {\rm AgNO_3} \, (aq)\! formula units to produce 1\; \rm mol \! of {\rm AgCl}\, (s)\! formula units.

Hence, it would take 0.313987\; \rm mol of {\rm AgNO_3} \, (aq)\! formula units to produce 0.313987\; \rm mol\! of {\rm AgCl}\, (s)\! formula units.

Calculate the volume of the {\rm AgNO_3} \, (aq)\! solution given that the concentration of the solution is 0.124\; \rm mol \cdot L^{-1}:

\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}.

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of {\rm AgNO_3} \, (aq)\!.)

In other words, approximately 2.53\; \rm L of that {\rm AgNO_3} \, (aq)\! solution would be required.

4 0
3 years ago
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