Answer:
a. 2nd order reaction.
b. The first step is the slow step.
Explanation:
r = k[NO][Cl₂]
a. The reaction is first-order in [NO] and first-order in [Cl₂], so it is second-order overall.
b. The first step is the slow step, because it predicts the correct rate law.
c. is wrong. Doubling [NO] would double the rate, because the reaction is first-order in [NO].
d. is wrong. Cutting [Cl₂] in half would halve the rate, because the reaction is first-order in [Cl₂].
e. is wrong. The molecularity is two, because two particles are colliding.
f. is wrong. Both steps are bimolecular.
<span>Answer: The 1 kg of water will reach the lowest temperature
</span>
Both of the objects is water so their specific heat should be same. The heat removed from the 2kg of water which is 2 times of mass than the 1kg water. Since the heat removed, both of their temperatures will drops. But 2kg water temperature drops will be half of 1kg water.
Answer:
6400 molecules / cm^3
Explanation:
10.6*10^-16 mol/L * 6.022*10^23 molecules/mol * 1 L / 1000 cm^3 * 1 / 100 L = 6400 molecules / cm^3
According to the information in the graph, it can be inferred that the amount of solute that will precipitate out of solution at 20°C is 130 grams.
<h3>How to calculate the amount of solute that precipitates out of solution?</h3>
To calculate the amount of solute that precipitates out of solution we must identify the solute data at 80°C and 20°C and identify the difference as shown below:
- Quantity of solute at 80°C: 170 grams.
- Quantity of solute at 20°C: 40 grams.
- 170 grams - 40 grams = 130 grams
According to the above, the amount of solute that will precipitate out of solution due to the change in temperature is 130 grams of KNO3.
Note: This question is incomplete because the graph is missing. Here is the graph
Learn more about solute in: brainly.com/question/7932885
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Answer:
550 m/s
Explanation:
The average molecular speed (v) is the speed associated with a group of molecules on average. We can calculate it using the following expression.

where,
We can use the info of argon to calculate the temperature for both samples.

Now, we can use the same expression to find the average molecular speed in a sample of Ne gas.
