Answer:
454.3 g.
Explanation:
1.0 mol of CaO liberates → – 64.8 kJ.
??? mol of CaO liberates → - 525 kJ.
∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.
<em>∴ The no. of grams of CaO needed = no. of moles x molar mass</em> = (8.1 mol)(56.077 g/mol) = <em>454.3 g.</em>
1.1214 mL will a 0.205-mole sample of He occupy at 3.00 atm and 200 K.
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Using equation PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.
Given data:
P= 3.00 atm
V= ?
n=0.205 mole
R= 
T=200 K
Putting value in the given equation:
V= 1.1214 mL
Learn more about the ideal gas here:
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Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.
Pretty sure it’s ,B. Microscope!
Answer:
H2C2O4.2H20 → CO2 + CO + H2O
Explanation:
Oxalic acid crystals are nothing but dehydrated oxalic acid (H2C2O4 . 2H2O).
On heating, the water of crystallization is lost first. Then, the dehydrated oxalic acid decomposes into carbon dioxide(CO2), carbon monoxide(CO) and water(H2O).
Equations involved :
H2C2O4 . 2H2O → H2C2O4 + 2H2O
H2C2O4 → CO2 + HCOOH (FORMIC ACID)
HCOOH → CO + H2O
Overall equation : H2C2O4.2H20 → CO2 + CO + H2O
