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katovenus [111]
3 years ago
10

Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.

Chemistry
1 answer:
weeeeeb [17]3 years ago
4 0

Answer:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2HBr(g)\rightarrow H_2(g)+Br_2(g)

Thus, the rate is given as:

rate=-\frac{1}{2} \frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [Br_2]}{\Delta t} =\frac{\Delta [H_2]}{\Delta t}

It is necessary to remember that each concentration to time interval is divided into the stoichiometric coefficient, that is why HBr has a 1/2. Moreover, the concentration HBr is negative since it is a reactant and it has a negative rate due to its consumption.

Therefore, the answer is:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Best regards.

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Silver metal can be prepared by reducing its nitrate, AgNO3 with copper according to the following equation:
Lerok [7]

%yield = 88.5%

<h3>Further explanation</h3>

Given

Reaction

Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)

Required

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Solution

mol AgNO₃(MW=169,87 g/mol) :

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= 127 : 169.87

= 0.748

mol Ag from equation :

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= 0.748

Mass Ag (theoretical) :

= mol x Ar Ag

= 0.748 x 108

= 80.784

% yield = (actual/theoretical) x 100%

%yield = 71.5/80.784 x 100%

<em>%yield = 88.5%</em>

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Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

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