Answer:
V₂ = 2.91 L
Explanation:
Given data:
Initial volume = 3.50 L
Initial temperature = 90.0°C (90+273 = 363 K)
Final temperature = 30.0 °C ( 30 +273 = 303 K)
Final volume = ?
Solution:
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
V₁/T₁ = V₂/T₂
3.50 L / 363 K) = V₂ / 303 K)
V₂ = 0.0096 L/K × 303 K
V₂ = 2.91 L
A: 2H₂ + O₂ → 2H₂O
H: 4 H:4
O: 2 O: 2
The equation is balanced.
B. 2S + 3O₂ → 2SO₃
S: 2 S: 2
O: 6 O: 6
The equation is balanced.
C. Li + Cl₂ → LiCl
Li: 1 Li: 1
Cl: 2 Cl: 1
The equation is not balanced.
2Li + Cl₂ → 2LiCl
Li: 2 Li: 2
Cl: 2 Cl: 2
D: 2K + 2H₂O → H₂ + 2KOH
K: 2 K: 2
H: 4 H: 4
O: 2 O: 2
The equation is balanced.
E: 2Fe + Cu(NO₃)₂ → 2Cu + Fe(NO₃)₂
Fe: 2 Fe: 1
Cu: 1 Cu: 2
N: 2 N: 2
O: 6 O: 6
The equation is not balanced.
2Fe + 2Cu(NO₃)₂ → 2Cu + 2Fe(NO₃)₂
Fe: 2 Fe: 2
Cu: 2 Cu: 2
N: 4 N: 4
O: 12 O: 12
The following equations that are balanced are A, B, and D.
Answer:
CO2 is a trigonal planar.
After ion exchange with zeolite in water filter to remove calcium by substitute natrium ion with calcium ion, the weight of zeolite become 1.9578 kg.
ion exchange calculation can be done by using basic stoichiometry if we know how the reaction of the ion exchange. In this case where zeolite with natrium ion will be exchanged with calcium. The reaction of exchange will be like this:
⇒ 
Calculate the molecule weight of both zeolite
Atomic weight data:
Na = 23 gram/mole
Si = 28 gram/mole
Al = 27 gram/mole
O = 16 gram/mole
Ca = 40 gram/mole
Molecule weight
= 142 gram/mole
= 278 gram/mole
Calculate mole from 2kg of zeolite with natrium :
2000 gram/ 142 = 14.0845 mole
Based on the stoichiometry, we got the mole of zeolite with calcium:
14.0845 mole x 1/2 = 7.0423 mole
Weight of zeolite after the ion exchanged with calcium:
7.0423 mole x 278 = 1957.8 gram = 1.9578 kg
Learn more about ion exchange here: brainly.com/question/28203551
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