Answer:
Potassium iodide increases the decomposition rate of hydrogen peroxide.
Explanation:
Potassium iodide increases the decomposition rate of hydrogen peroxide because potassium iodide act as a catalyst. A catalyst speed up the process of chemical reaction without reacting with the molecules present in reaction. If the potassium iodide is not present as a catalyst for the decomposition of hydrogen peroxide then the decomposition of hydrogen peroxide takes too much time because the catalyst is absent that speed up the reaction.
Increases with
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Answer:
A. 30cm³
Explanation:
Based on the chemical reaction:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
<em>1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂</em>
<em />
To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:
<em>Moles CaCO₃ -Molar mass: 100.09g/mol-</em>
1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles
<em>Moles HCl:</em>
50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles
For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:
2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.
The moles produced of CO₂ are:
2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂
Using PV = nRT
<em>Where P is pressure = 1atm assuming STP</em>
<em>V volume in L</em>
<em>n moles = 1.25x10⁻³ moles CO₂</em>
<em>R gas constant = 0.082atmL/molK</em>
<em>T = 273.15K at STP</em>
<em />
V = nRT / P
1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V
0.028L = V
28cm³ = V
As 28cm³ ≈ 30cm³
Right option is:
<h3>A. 30cm³</h3>
As we know that 760 mmHg is equal to 1 atm.
So,
If 760 mmHg is equal to = 1 atm
Then
738 mmHg will be equal to = X atm
Solving for X,
X = (738 mmHg × 1 atm) ÷ 760 mmHg
X = 0.971 atm
Result:
738 mmHg is equal to 0.971 atm.
Answer:
C
Explanation:
Since the solution have an observable color, that means that it absorbs light in the visible region hence it can be determined by colorimetry. Secondly, KMnO4 is a reducing agent which can be titrated against an oxidizing agent and it's concentration accurately determined.
