3I₂ + 2Al → 2AlI₃
m(I₂)=3M(I₂)m(Al)/{2M(Al)}
m(I₂)=3*253.8*20.4/{2*27.0}=287.64 g
a resource that cannot be replenished in a short period of time
Answer:
5SiO2 + 2CaC2 = 5Si + 2CaO + 4CO2
Explanation:
balancing equations is a lot of trial and error. My strategy to approaching this equation was to get the O's balanced. After trying several combonations I found that I needed 10 O's on each side of the equation for the other elements to match up. After I balanced the O's, I balanced my C's to 4 on each side. Then I balanced my Ca's to have 2 on each side. And last but not least I balanced my Si to have 5 on each side.
Answer:

Explanation:
Given:
Pressure = 745 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 745 / 760 = 0.9803 atm
Temperature = 19 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (19 + 273.15) K = 292.15 K
Volume = 0.200 L
Using ideal gas equation as:

where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K
⇒n = 0.008174 moles
From the reaction shown below:-

1 mole of
react with 2 moles of 
0.008174 mole of
react with 2*0.008174 moles of 
Moles of
= 0.016348 moles
Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)
So,



Answer:
Only two elements are liquid at standard conditions for temperature and pressure: mercury and bromine. Four more elements have melting points slightly above room temperature: francium, caesium, gallium and rubidium.
Explanation: