Density = Mass / Volume
V = 1.00 * 4.00 * 2.50 = 10 cm³
22.57 g/cm³ = Mass / 10 cm³
M = 22.57 g/cm³ * 10 cm³
M = 225.7 g
Answer: The mass of the block of osmium is 225.7 g.
Hey there!
* Converts 1750 dm³ in liters :
1 dm³ = 1 L so 1750 dm³ = 1750 liters
* Convertes 125,000 Pa in atm :
1 Pa = 9.86*10⁻⁶ atm so 9.86*10⁻⁶ / 125,000 => 1.233 atm
* Convertes 127ºC in K :
127 + 273.15 => 400.15 K
R = 0.082 atm.L/mol.K
Finally, it uses an equation of clapeyron :
p * V = n * R * T
1.233 * 1750 = n * 0.082 * 400.15
2157.75 = n * 32.8123
n = 2157.75 / 32.8123
n = 65.76 moles
hope this helps!
Answer:
Any of the answers given will work
Explanation:
I literally just did it.
At the end of the reaction, the catalyst is UNCHANGED.
:)
Answer:
Silver Acetate would be the Limiting Reagent.
Explanation:
The balance chemical equation for the given double displacement reaction is as;
HCl + AgC₂H₃O₂ → AgCl + HC₂H₃O₂
Step 1: <u>Calculate Moles of Starting Materials:</u>
Moles of HCl:
Moles = Mass / M.Mass
Moles = 72.9 g / 36.46
Moles = 1.99 moles
Moles of AgC₂H₃O₂:
Moles = 150 g / 166.91 g/mol
Moles = 0.898 moles
Step 2: <u>Find out Limiting reagent as:</u>
According to balance chemical equation.
1 mole of HCl reacts with = 1 mole of AgC₂H₃O₂
So,
1.99 moles of HCl will react with = X moles of AgC₂H₃O₂
Solving for X,
X = 1.99 mol × 1 mol / 1 mol
X = 1.99 mol of AgC₂H₃O₂
Hence, to completely consume 1.99 moles of Hydrochloric acid we will require 1.99 moles of Silver Acetate, But, we are provided with only 0.898 moles of Silver Acetate. This means Silver Acetate will consume first in the reaction therefore, it is the LIMITING REAGENT.
