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CaHeK987 [17]
3 years ago
15

Catalytic converters made of palladium (Pd) reduce automobile pollution by catalyzing the reaction between unburned hydrocarbons

and oxygen. How does Pd increase the rate of this reaction?
A: By cooling the reactants
B: By splitting the oxygen atoms
C: By giving the hydrocarbons a negative charge
D: By decreasing the activation energy

Chemistry
2 answers:
Lubov Fominskaja [6]3 years ago
5 0

Answer: D: By decreasing the activation energy

Explanation: A catalyst is a substance which increases the rate of a reaction by taking the reaction through a different path which involves lower activation energy and thus more molecules can cross the energy barrier and convert to products.

Activation energy is the extra energy that must be supplied to reactants in order to cross the energy barrier and thus convert to products.

The catalyst itself does not take part in the chemical reaction and is regenerated as such at the end.

Setler79 [48]3 years ago
3 0
Almost all catalysts work by lowering the activation energy of the reaction with no change in the free energy of the reaction 

- So in this case we can say that palladium  reduce automobile pollution by catalyzing the reaction between un-burned hydrocarbons and oxygen :

D. by decreasing the activation energy 
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Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature
weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}     ------>NO_{(g)}

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

\delta H^0__{R,T_2} = \delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'

where:

\delta H^0__{R} = enthalpy of reaction

{\delta C_p(T')} = the difference in the heat capacities of the products and the reactants.

∴

\delta H^0__{R,435K} = \delta H^0__{R,298.15K} + \int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'

= 1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

\delta H^0__{R,435K} ≅ 91383 J

6 0
3 years ago
WILL GIVE 50 POINTS FOR ALL ANSWERS!!! PLEASE HELP!
blondinia [14]

Answer:

the answer would be (A.) and (D.).

Explanation:

the reason for that being is because if calcium sulfate is a main component of plaster of  paris you would need to find out what is in it that makes it the main component aka (the formula) therefore part of the answer is (A.). The other part of the answer was (D.) because you would need to find the amount of calcium sulfate that contain 12 grams of oxegeon atoms because you finding the answer to that could lead to the answer of what is the main component of plaster of Paris.

3 0
3 years ago
If the volume of the original sample in Part A ( P1 = 242 torr , V1 = 27.0 L ) changes to 80.0 L , without a change in the tempe
Brrunno [24]

The new pressure is 81.675 torr

Since temperature and moles are held constant, we use Boyle's Law:

A gas law known as Boyle's law asserts that a gas's pressure is inversely proportional to its volume when it is held at a fixed temperature and of a given mass.

To put it another way, as long as the temperature and volume of the gas remain constant, the pressure and volume of the gas are inversely proportional to one another.

The Anglo-Irish chemist Robert Boyle proposed Boyle's law in the year 1662.

P1V1=P2V2. Simply plug in your values. The units can remain in torr. Converting to atmospheres is not needed.

(242 torr)(27.0 L)=P2(80.0 L)

P2=[(242)(27)]/80 = 81.675 torr

Hence The new pressure is 81.675 torr

Learn more about Boyle's Law here

brainly.com/question/26040104

#SPJ4

3 0
1 year ago
What is the [OH-] of a substance that has a pH of 11?
Deffense [45]

Answer:

0.001 M OH-

Explanation:

[OH-] = 10^-pOH, so

pOH + pH = 14 and 14 - pH = pOH

14 - 11 = 3

[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-

6 0
2 years ago
What is the mole fraction of methanol in a solution that contains 6.0 mol of methanol and 3.0 mol of water? The formula for meth
Ivenika [448]
The mole fraction of a product is the number of moles of the product divided by the total number of moles of the solution.

Here moles of methanol = 6.0 moles

Moles of solution = 6.0 moles of methanol + 3.0 moles of water = 9.0 moles of solution

Mole fraction of methanol = 6.0 / 9.0 = 0.67

Answer: 0.67  
4 0
3 years ago
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