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3 years ago

5

True or false? Recycling metal reduces fuel costs as less metal ore needs to be mined and transported.

1 answer:

chubhunter [2.5K]3 years ago

4 0

Answer:

True

Explanation:

True.

Recycling of metal means that there is a reduction in the cost of fuel, since much lesser metal ore would have to be mined and eventually transported. Recycling of metal also means saving of energy since we won't be mining ores from the beginning all over again, as compared. There is lesser number of quarries needed as well.

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Hydrogen reacts with chlorine to form hydrogen chloride (HCl (g), mc012-1.jpgHf = –92.3 kJ/mol) according to the reaction below.

Kobotan [32]

<u>Given:</u>

H2(g) + Cl2 (g) → 2HCl (g)

<u>To determine:</u>

The enthalpy of the reaction and whether it is endo or exothermic

<u>Explanation:</u>

Enthalpy of a reaction is given by the difference between the enthalpy of formation of reactants and products

ΔH = ∑nHf (products) - ∑nHf (reactants)

= [2Hf(HCl)] - [Hf(H2) + Hf(Cl2)] = 2 (-92.3) kJ = - 184.6 kJ

Since the reaction enthalpy is negative, the reaction is exothermic

<u>Ans:</u> The enthalpy of reaction is -184. kJ and the reaction is exothermic

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4 years ago

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

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Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

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