The number of moles of argon that must be released in order to drop.
Solution:
Initial Temperature = 25°c = 298 K
Final Temperature =125 °c = 398 K
Initial Moles (n1) = 0.40 mole
Now, Using the ideal gas law,
n1T1 = n2T2
0.400×298 = n2 × 398
n2 = 0.299 mol
Moles of Argon released
= 0.400-0.299
= 0.100 mol.
Pressure and force are related. That is using the physical equations if you know the other, you can calculate one using pressure = force/area. This pressure can be reported in pounds per square inch, psi, or Newtons per square meter N/m2. Kinetic energy causes air molecules to move faster. They hit the walls of the container more often and with greater force. The increased pressure inside the can may exceed the strength of the can and cause an explosion.
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Answer:
B
Explanation:
so that way they can hide away from the predators
It is elastic potential energy
Answer:
The answer to your question is below
Explanation:
Data
mass of CaCO₃ = 155 g
mass of HCl = 250 g
mass of CaCl₂ = 142 g
reactants = CaCO₃ + HCl
products = CaCl₂ + CO₂ + H₂O
1.- Balanced chemical reaction
CaCO₃ + 2HCl ⇒ CaCl₂ + CO₂ + H₂O
2.- Limiting reactant
molar mass of CaCO₃ = 40 + 12 + 48 = 100 g
molar mass of HCl = 2[1 + 35.5 ] = 73 g
theoretical proportion CaCO₃ /HCl = 100 / 73 = 1.37
experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62
As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃
3.-
Calculate the molar mass of CaCl₂
CaCl₂ = 40 + 71 = 111 g
100 g of CaCO₃ ------------------ 111 g of CaCl₂
155 g of CaCO₃ ----------------- x
x = (155 x 111) / 100
x = 17205 / 100
x = 172.05 g of CaCl₂
4.- percent yield
Percent yield = 142 / 172.05 x 100 = 82.5 %
5.- Excess reactant
100 g of CaCO₃ -------------------- 73 g of HCl
155 g of caCO₃ ------------------- x
x = (155 x 73)/100
x = 133.15 g
Mass of HCl = 250 - 133.15
= 136.9 g
