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chubhunter [2.5K]
3 years ago
7

The law of mass conservationis another fundamental law in chemistry. This law states that matter can neither be created nor dest

royed, which means that in a chemical reaction where all of the reactants is converted into products, the mass of reactant that reacts must be exactly equal to the mass of product produced. Next week you are going totest the law of mass conservation. To do this you will need to design your own experiment
Chemistry
1 answer:
kupik [55]3 years ago
4 0

Answer:

trukugjgfcliud

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The rate of formation of C in the reaction 2 A + B →2 C + 3 D is 2.7 mol dm−3 s −1 . State the reaction rate, and the rates of f
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Answer:

Rate of reaction =1.35mol.dm^{-3}.s^{-1}

Rate of consumption of A = 2.7mol.dm^{-3}.s^{-1}

Rate of consumption of B = 1.35mol.dm^{-3}.s^{-1}

Rate of formation of D = 4.15mol.dm^{-3}.s^{-1}

Explanation:

According to laws of mass action for the given reaction,

Rate= -\frac{1}{2}\frac{\Delta [A]}{\Delta t}=-\frac{\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}=\frac{1}{3}\frac{\Delta [D]}{\Delta t}

where, -\frac{\Delta [A]}{\Delta t} is rate of consumption of A, -\frac{\Delta [B]}{\Delta t} is rate of consumption of B, \frac{\Delta [C]}{\Delta t} is rate of formation of C and \frac{\Delta [D]}{\Delta t} is rate of formation of D

Here \frac{\Delta [C]}{\Delta t}=2.7mol.dm^{-3}.s^{-1}

So, Rate of reaction = (\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}

Rate of formation of D = (\frac{3}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{3}{2}\times 2.7mol.dm^{-3}.s^{-1})=4.15mol.dm^{-3}.s^{-1}

Rate of consumption of A = (\frac{2}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{2}{2}\times 2.7mol.dm^{-3}.s^{-1})=2.7mol.dm^{-3}.s^{-1}

Rate of consumption of B = (\frac{1}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}

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