Glucose and a plants and ur welcome
Answer:
The right answer is:
Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 1.0 gram of lead (IV) oxide
Explanation:
Based on the given information this reaction is the catalytic decomposition of H₂O₂ into water and oxygen using Lead (IV) oxide as a catalyst.
- The catalyst surface area is directly proportional to the reaction rate
- So, Replacing the powdered lead oxide with its large crystals would decrease the reaction rate due to the has larger surface area than its large crystals.
2. Also, Removing lead (IV) oxide from the reaction mixture the reaction rate decreased because as the catalyst is removed.
3. Using 50 cm³ of hydrogen peroxide doesn't affect the rate because the concentration of the reactant doesn't change.
4. Using 1.0 gram of lead (IV) oxide would decrease the reaction rate because the amount of catalyst decreased
So, The right answer is:
Replacing the powdered lead oxide with its large crystals
Removing lead (IV) oxide from the reaction mixture
Using 1.0 gram of lead (IV) oxide
<h2>
Answer:</h2>
44.06 g/mol
<h3>
Explanation:</h3>
We are given;
- Number of moles of unidentified gas as 1.674×10^-4 mol
- Time of effusion of unidentified gas 86.6 s
- Number of moles of Argon gas as 1.715×10^-4 mol
- Time of effusion of Argon gas is 84.5 s
We are supposed to calculate the molar mass of unidentified gas
<h3>Step 1: Calculate the effusion rates of each gas</h3>
Effusion rate = Number of moles/time
Effusion rate of unidentified gas (R₁)
= 1.674×10^-4 mol ÷ 86.6 s
= 1.933 × 10^-6 mol/s
Effusion rate of Argon gas (R₂)
= 1.715×10^-4 mol ÷ 84.5 sec
= 2.030 × 10^-6 mol/s
<h3>Step 2: Calculate the molar mass of unidentified gas</h3>
- Assuming the molar mass of unidentified gas is x;
- We can use the Graham's law of effusion to find x;
- According to Graham's law of diffusion;
But, Molar mass of Argon is 39.948 g/mol
Therefore;
Solving for X
x = 44.06 g/mol
Therefore, the molar mass of the identified gas is 44.06 g/mol
Answer:
sigma bond and represent two electrons
The following chemical reaction will occur:
Br₂ (l) + 2 NaI (s) → 2 NaBr (s) + I₂ (s)
Explanation:
Because the bromide (Br₂) have a higher reactivity than iodide (I₂) it is able the remove the iodide from its salts. So the bromide will react with sodium iodine (NaI) to produce sodium bromide (NaBr) and iodine.
The chemical reaction is:
Br₂ (l) + 2 NaI (s) → 2 NaBr (s) + I₂ (s)
where:
(l) - liquid
(s) - solid
Learn more about:
balancing chemical equations
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