1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm
Answer: Non polar solvents
Explanation:
Since with increasing the size of alkyl group hydrophobic nature increases and solubility in polar solvents decreases .
Hence Carboxylic acids with more than 10 carbon atoms, solubility is more in non polar solvents.
Answer:
124.91mL
Explanation:
Given parameters:
P₁ = 1.08atm
V₁ = 250mL
T₁ = 24°C
P₂ = 2.25atm
T₂ = 37.2°C
V₂ = ?
Solution:
To solve this problem, we are going to apply the combined gas law;
P, V and T represents pressure, volume and temperature
1 and 2 delineates initial and final states
Convert the temperature to kelvin;
T₁ = 24°C, T₁ = 24 + 273 = 297K
T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K
Input the variables and solve for V₂
V₂ = 124.91mL
Answer: CuI₂ + Br₂
Explanation:
1) The activity series F > Cl > Br > I means that F is the most active and I is the least active of those four elements (the halogens, group 17 in the periodic table).
The activity is a measure of how eager is an element to react compared to other elements in the series in a single replacement reaction.
2) Choice 1: CuI₂ + Br₂
Since the activity of Br is higher than that of I, Br will react with CuI₂, displacing I, which will be left alone, as per this chemical equation:
CuI₂ + Br₂ → CuBr₂ + I₂
Being I less active than Br, it cannot displace Br in CuBr₂.
3) Choice 2: Cl₂ + AlF₃
Being Cl less active than F, the former will not displace the latter, and the reaction will not proceed.
4) Choice 3: Br₂ + NaCl
Again, being Br less active than Cl, the former will not displace the latter, and the reaction will not proceed.
5) Choice 4: CuF₂ + I₂
Once more, being I less active than F, the former will not displace the latter, and the reaction will not proceed.
