Answer:
volume of the container will decreases if pressure increases.
Explanation:
According to Boyle's law:
Pressure is inversely proportional to volume which means if pressure of a gas increases the volume of the gas will decreases as gas molecules will collide and come closer forcefully so volume will decreases. And its formula for determining volume and pressure is:
<em>PV=nRT</em>
where "R" is a ideal gas constant
"T" is temperature and
"n" is number of particles given in moles while "V" is volume and "P" is pressure.
Answer:
1.8 moles of O₂
Explanation:
The balance chemical equation for said double replacement (photosynthesis) reaction is as follow;
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
According to balance chemical equation,
6 moles of O₂ are produced by = 6 moles of CO₂
So,
1.8 moles of O₂ will be produced by = X moles of O₂
Solving for X,
X = 1.8 mol × 6 mol / 6 mol
X = 1.8 moles of O₂
Stoichiometric problems in which moles are given and moles or other reactant or product asked are the simplest problems. One should only write the balanced chemical equation and perform above method to find the required moles.
C) Fluorine is the most reactive among halogens
Answer:
148 g
Explanation:
Step 1: Write the balanced equation for the decomposition of sodium azide
2 NaN₃ ⇒ 2 Na + 3 N₂
Step 2: Calculate the moles corresponding to 95.8 g of N₂
The molar mass of N₂ is 28.01 g/mol.
95.8 g × 1 mol/28.01 g = 3.42 mol
Step 3: Calculate the moles of NaN₃ needed to form 3.42 moles of N₂
The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.42 mol = 2.28 mol.
Step 4: Calculate the mass corresponding to 2.28 moles of NaN₃
The molar mass of NaN₃ is 65.01 g/mol.
2.28 mol × 65.01 g/mol = 148 g
Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g
