Answer:
5.5 L
Explanation:
First we <u>convert 10 g of propane gas</u> (C₃H₈) to moles, using its <em>molar mass</em>:
- 10 g ÷ 44 g/mol = 0.23 mol
Then we <u>use the PV=nRT formula</u>, where:
- P = 1 atm & T = 293 K (This are normal conditions of T and P)
- R = 0.082 atm·L·mol⁻¹·K⁻¹
1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K
Answer:
M of Al=33.09g or 0.0331kg
Explanation:
Heat Energy= specific heat*mass*change in temperature
H=M*C*T
make M subject of the formula
M=H/CT
M=685J/0.90J/g°C*(45°C-22°C)
M=685J/0.90J/g°C*23°C
M=685J/20.7J/g
M=33.09g or 0.0331kg
Answer:
The final product of the reaction is (<em>2S,3S</em>)-2-ethoxy-3-methylpentane.
Explanation:
The given reaction undergoes
mechanism in which the nucleophile attacks the backside and it is substituted by the elimination of bromine.
Due to the backside attack of nucleophile , the inverse in stereo-chemistry is observed.
After the substitution of ethoxy group, the configuration is assigned according to the priority it shows clock wise direction(R) - configuration.
When hydrogen faces the front side , it results shows inverse configuration i.e, S- configuration.
The chemical reaction is as follows.
<span>The outer electrons are not as tightly bound as ones closer to the nucleus</span>
32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is
of butane reacted
Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol
= 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol
= 99.3 g of CO₂
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced