Answer:
Explanation:
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In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:
Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:
Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:
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That is called one centimeter
Answer:
%age Yield = 34.21 %
Explanation:
The balance chemical equation for the decomposition of KClO₃ is as follow;
3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O
Step 1: Calculate moles of H₃PO₄ as;
Moles = Mass / M/Mass
Moles = 334.6 g / 97.99 g/mol
Moles = 3.414 moles
Step 2: Find moles of K₃PO₄ as;
According to equation,
1 moles of H₃PO₄ produces = 1 moles of K₃PO₄
So,
3.414 moles of H₃PO₄ will produce = X moles of K₃PO₄
Solving for X,
X = 1 mol × 3.414 mol / 1 mol
X = 3.414 mol of K₃PO₄
Step 3: Calculate Theoretical yield of K₃PO₄ as,
Mass = Moles × M.Mass
Mass = 3.414 mol × 212.26 g/mol
Mass = 724.79 g of K₃PO₄
Also,
%age Yield = Actual Yield / Theoretical Yield × 100
%age Yield = 248 g / 724.79 × 100
%age Yield = 34.21 %
Answer:
Redox type
Explanation:
The reaction is:
2Cr + 3Fe(NO₃)₂ → 2Fe + 2Cr(NO₃)₃
2 moles of chromium can react to 3 moles of iron (II) nitrate in order to produce 2 moles of iron and 2 moles of chromium nitrate.
If we see oxidation state, we see that chromium changes from 0 to +3
Iron changed the oxidation state from +2 to 0
Remember that elements at ground state has 0, as oxidation state.
Iron is being reduced while chromium is oxidized. Then, the half reactions are:
Fe²⁺ + 2e⁻ ⇄ Fe (Reduction)
Cr ⇄ Cr³⁺ + 3e⁻ (Oxidation)
When an element is being reduced, while another is being oxidized, we are in prescence of a redox reaction.
Answer:
Explanation:
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In this case, we can identify the solution to this problem via the Dalton's rule because the partial pressure of helium is given by:
Whereas the mole fraction of helium is calculated by firstly obtaining the moles and then the mole fraction:
Then, we calculate the partial pressure as shown below:
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