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3 years ago

Chem 2"

Chemistry

1 answer:

svet-max [94.6K]3 years ago

8 0

b.It is a heterogeneous catalyst, so only a small amount is needed and it is easily separated from the products.

Explanation:

Platinum is best described as a heterogeneous catalyst in which only a small amount of it is needed and it can be easily separated from products.

A heterogeneous catalyst is a catalyst that exists in a different phase than the reactants.

A homogeneous catalyst is one that exists in the same phase as the reactants.

A catalyst is a substance that speeds up the rate of chemical reactions.
They help to lower the activation barrier for the reaction and are not used up.
The reactants in the car exhausts are in the gaseous phases.
Platinum is a solid metal.
Therefore we can see that it is a heterogeneous catalyst.

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Biological catalyst brainly.com/question/8413755

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The reaction :

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$Cr_2O_7^{2-}(aq)\rightarrow 2Cr^{3+}(aq)+7H_2O(l)$

Step 3: Balance the hydrogen atom by adding hydrogen ions on the side where water is absent:

$Cr_2O_7^{2-}(aq)+14H^+(aq)\rightarrow 2Cr^{3+}(aq)+7H_2O(l)$

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$Cr_2O_7^{2-}(aq)+14H^+(aq)+6e^-\rightarrow 2Cr^{3+}(aq)+7H_2O(l)$

The balanced half-reaction for the reduction of dichromate ion to chromium ion in acidic aqueous solution:

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Hot lead with a mass of 200.0 g of (Specific heat of Pb = 0.129 J/gºC) at 176.4°C was dropped into a calorimeter containing an u

Morgarella [4.7K]

the Calorimetry relationships you can find the amount of water in the calorimeter is m = 21.3 g

given parameters

Lead mass M = 200.0 g
Initial lead temperature T₁ = 176.4ºC
Specific heat of Lead $c_{e Pb}$ = 0.129 J / g ºC
Sspecific heat of water $c_{e H_2O}$ = 4.186 J / g ºC
Initial water temperature T₀ = 21.7ºC
Equilibrium temperature T_f = 56.4ºC

to find

The body of water

Thermal energy is the energy stored in the body that can be transferred as heat when two or more bodies are in contact. Calorimetry is a technique where the energy is transferred between the body only in the form of heat and in this case the thermal energy of the lead is transferred to the calorimeter that reaches the equilibrium that the thematic energy of the two is equal

Q_{ceded} = Q_{absorbed}

Lead, because it is hotter, gives up energy

Q_{ceded} = M c_{e Pb} (T₁ - T_f)

The calorimeter that is colder absorbs the heat

Q_{absrobed} = m c_{e H_2O} (T_f - T₀)

where M and m are the mass of lead and water, respectively, c are the specific heats, T₁ is the temperature of the hot lead, T₀ the temperature of cold water and T_f the equilibrium temperature

M c_{ePb} (T₁ - T_f) = m c_{eH2O} (T_f - T₀)

m = $\frac{ M\ c_{ePb} \ (T_1 - T_f)}{c_{eH_2O} \ (T_f - T_o)}$

let's calculate

m = $\frac{200 \ 0.129 (176.4-56.4)}{ 4.186 \ (56.4 -21.7)}$

m = 3096 / 145.25

m = 21.3 g

Using the Calorimetry relationships you can find the amount of water in the calorimeter is:

m = 21.3 g

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