Answer:
The moles of sucrose that are available for this reaction is 0.0292 moles
Explanation:
Combustion is an specifyc reaction where the reactants react with O₂ in order to produce CO₂ and H₂O
This combustion is: C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
We have to conver the mass to moles, to find out the limiting reactant
10 g . 1 mol / 342 g = 0.0292 moles of sucrose
8 g . 1mol / 32g = 0.250 moles of O₂
The moles of sucrose that are available for this reaction is 0.0292 moles
Before we start to work with the equation we must find the limiting reactant. When you find it, you can do all the calculations.
Answer:
<em>Nitrogen = 17.07%</em>
Explanation:
We are asked to calculate the percentage by mass of nitrogen in Ca(NO3)2.
The molar mass of calcium nitrate is 164,088 g/mol.
In this 164.088 g/mol there are 2 nitrogen atoms.
Molar mass of nitrogen = 14,0067 g/mol
% Nitrogen = molar mass of N/total molar mass of the compound
% Nitrogen =[(2 x 14.0067)/164.088]*100
% Nitrogen = 17.07%
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Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
Answer: (2) decreasing the concentration of HCl(aq) to 0.1 M
Explanation: Rate of a reaction depends on following factors:
1. Size of the solute particles: If the reactant molecules are present in smaller size, surface of particles and decreasing the size increases the surface area of the solute particles. Hence, increasing the rate of a reaction.
2. Reactant concentration: The rate of the reaction is directly proportional to the concentration of reactants.
3. Temperature: Increasing the temperature increases the energy of the molecules and thus more molecules can react to give products and rate increases.
(1) Increasing the initial temperature to 25°C will increase the reaction rate.
(2) Decreasing the concentration of HCl(aq) to 0.1 M will decrease the reaction rate due to lesser concentration.
(3) Using 1.2 g of powdered Mg will increase the reaction rate due to large surface area.
(4) Using 2.4 g of Mg ribbon will increase the reaction rate due to high concentration of reactants.
Answer:
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