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Veronika [31]
4 years ago
5

Platinum is used to catalyze the hydrogenation of ethylene: H2(g)+CH2CH2(g)−⟶Pt(s)CH3CH3(g)H2(g)+CH2CH2(g)−⟶Pt(s)CH3CH3(g) Chlor

ofluorocarbons (CFCs) catalyze the conversion of ozone (O3O3) to oxygen gas (O2O2): 2O3(g)−⟶CFC(g)3O2(g)2O3(g)−⟶CFC(g)3O2(g) Magnesium catalyzes the disproportionation of hydrogen peroxide to produce water and oxygen: 2H2O2(aq)−⟶Mg(s)2H2O(l)+O2(g)2H2O2(aq)−⟶Mg(s)2H2O(l)+O2(g) What type of catalysts are platinum, CFCs, and magnesium under these conditions? Drag the appropriate items to their respective bins.
Chemistry
1 answer:
Dafna1 [17]4 years ago
4 0

Explanation:

Homogeneous catalysis refers to catalytic reactions where the catalyst is in the same phase as the reactants.

Heterogeneous catalysis is the type of catalysis where the phase of the catalyst differs from the phase of the reactants or products.

Platinum is used to catalyze the hydrogenation of ethylene:

H2(g)+CH2CH2(g)− ⟶ Pt(s) CH3CH3(g)

In this reaction, the platinum is in the solid state. While the other species (reactants and products) are in their gaseous state.

This reaction is Heterogenous catalysis.

Chlorofluorocarbons (CFCs) catalyze the conversion of ozone (O3O3) to oxygen gas (O2O2):

2O3(g)− ⟶ CFC(g) 3O2(g)

The catalyst is in the same gaseous state as the reactant and product.

This reaction is Homogenous catalysis.

Magnesium catalyzes the disproportionation of hydrogen peroxide to produce water and oxygen:

2H2O2(aq)− ⟶ Mg(s) 2H2O(l) + O2(g)

In this reaction, the Magnesium is in the solid state. While the other species (reactants and products) are in their gaseous state.

This reaction is Heterogenous catalysis.

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Quantitative

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Qualitative

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3 0
3 years ago
A red sports drink contains Red 40 dye. A 5.4 mL aliquot of this sports drink was diluted to 25.0 mL with deionized water in a v
miskamm [114]

Answer:

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

Explanation:

Concentration of red dye in sport drink before dilution C_1=?

Volume of the sport drink before dilution V_1=5.4 mL

Concentration of red dye in sport drink after dilution C_2=18.1 ppm

Volume of the sport drink after dilution V_2=25.0 mL

C_1V_1=C_2V_2( dilution )

C_1=\frac{C_2V_2}{V_1}=\frac{18.1 ppm\times 25.0 mL}{5.4 mL}

C_1=83.80 ppm\approx 84.0ppm

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

6 0
3 years ago
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