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svlad2 [7]
3 years ago
10

Why are tin granules used instead of just a piece of tin?

Chemistry
2 answers:
Soloha48 [4]3 years ago
4 0

The reason why Tin granules is mostly used or preferred than just a piece of Tin because a piece of Tin is plainly metal, in which this will be difficult in having it to be oxidized when it is exposed or in the air rather than Tin granules.

VikaD [51]3 years ago
4 0

Answer:

To increase the surface area of reaction.

Explanation:

The tiny granules are used as they increase the surface-area to volume ratio. Smaller granules have a higher surface area to volume ratio. In a chemical reaction, it means that the reaction increases as there is more particle to particle reaction.

Hence the granules are preferred to solids.

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You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
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8 0
3 years ago
Consider the following reaction where Kc = 1.80×10-2 at 698 K:
Klio2033 [76]

Answer:

The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.

Explanation:

The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.

For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:

Qc=\frac{[C]^{c}*[D]^{d}  } {[A]^{a}*[B]^{b}}

In this case:

Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}

Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:

  • [H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}=2.09*10⁻² \frac{moles}{liter}
  • [I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}=4.14*10⁻² \frac{moles}{liter}
  • [I_{2} ]=\frac{0.280 moles}{1 Liter}= 0.280 \frac{moles}{liter}

So,

Qc=\frac{2.09*10^{-2} *4.14*10^{-2}  } {0.280^{2} }

Qc= 0.011

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. <u><em>The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.</em></u>

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