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4 years ago

OMG SOMEONE PLEASE HELP I DONT KNOW HOW TO DO THIS

Chemistry

1 answer:

uranmaximum [27]4 years ago

4 0

3-25200 m
4- 0.0023 cl
5-
6- 23 mm
7-4500mm
8-450000ml
9-0.45kg
10-

Sorry I couldn’t get them all but there is most of the answers

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Calculate ΔHrxn for the following reaction: Fe₂O₃(s)+3CO(g)→2Fe(s)+3CO₂(g) Use the following reactions and given ΔH′s. 2Fe(s)+3/

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Answer : The enthalpy change of reaction is -23.9 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given final reaction is,

$Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $2Fe(s)+\frac{3}{2}O_2(g)\rightarrow Fe_2O_3(s)$ $\Delta H^o_1=-824.2kJ$

(2) $CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$ $\Delta H^o_2=-282.7kJ$

First we will reverse the reaction 1 and multiply equation 2 by 3 then adding both the equation, we get :

(1) $Fe_2O_3(s)\rightarrow 2Fe(s)+\frac{3}{2}O_2(g)$ $\Delta H^o_1=+824.2kJ$

(2) $3CO(g)+\frac{3}{2}O_2(g)\rightarrow 3CO_2(g)$ $\Delta H^o_2=3\times (-282.7kJ)=-848.1kJ$

The expression for final enthalpy is,

$\Delta H=\Delta H^o_1+\Delta H^o_2$

$\Delta H=(+824.2kJ)+(-848.1kJ)$

$\Delta H=-23.9kJ$

Therefore, the enthalpy change of reaction is -23.9 kJ

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The protein catalase catalyzes the reaction The Malcolm Baldrige National Quality Award aims to:

Marina CMI [18]

The question is missing a part, so the complete question is as follows:

The protein catalase catalyzes the reaction The Malcolm Bladrigde National Quality Awards aims to: 2H2O2 (aq) ⟶ 2H2O (l) + O2 (g) and has a Michaelis-Menten constant of KM = 25mM and a turnover number of 4.0 × 10 7 s -1. The total enzyme concentration is 0.012 μM and the intial substrate concentration is 5.14 μM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Calculate the initial rate, R (often written as V0), of this reaction.

1) Calculate Rmax

The turnover number (Kcat) is a ratio of how many molecules of substrate can be converted into product per catalytic site of a given concentration of enzyme per unit of time:

Kcat = $\frac{Vmax}{Et}$ ,

where:

Vmax is maximum rate of reaction when all the enzyme sites are saturated with substrate

Et is total enzyme concentration or concentration of total enzyme catalytic sites.

Calculating:

Kcat = $\frac{Vmax}{Et}$

Vmax = Kcat · Et

Vmax = 4× $10^{7}$ · 1.2 × $10^{-8}$

Vmax = 4.8 × $10^{-1}$ M

2) Calculate the initial rate of this reaction (R):

The Michaelis-Menten equation studies the dynamics of an enzymatic reaction. This model can explain how an enzyme enhances the rate of a reaction and how the reaction rate depends on the concentration of the enzyme and its substrate. The equation is:

V0 = $\frac{[S].(Vmax)}{KM + [S]}$ , where:

[S] is the substrate's concentration

KM is the Michaelis-Menten constant

Substituting [S] = 5.14 × $10^{-6}$ , KM = 2.5 × $10^{-4}$ and Vmax = 4.8 × $10^{-1}$ , the result is V0 = 0.478 M.

The answers are Vmax = 4.8 × $10^{-1}$ M and V0 = 0.478 M.

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