Answer:
Concentration of product at equilibrium ;
![[H^+]=0.0000229 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.0000229%20M)
![[CN^-]=0.0000229 M](https://tex.z-dn.net/?f=%5BCN%5E-%5D%3D0.0000229%20M)
Explanation:
initially
0.85 M 0 0
(0.85-x)M x x
The equilibrium constant of reaction = 
The expression of an equilibrium cannot can be written as:
![K_c=\frac{[H^+][CN^-]}{[HCN]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH%5E%2B%5D%5BCN%5E-%5D%7D%7B%5BHCN%5D%7D)
Solving for x:
x = 0.0000229
Concentration of product at equilibrium ;
A compound with the formula C6H12 is not considered a Saturated hydrocarbon.
Why is C6H12 isn't considered a Saturated hydrocarbon?
The ring's presence demonstrates that it is unsaturated. Keep in mind that the general formula for aliphatic hydrocarbons, CnH2n+2, serves as the foundation for its saturation. A chemical is unsaturated if it does not meet this requirement.
Example:
Hexane (C6H14)
C = 6; H = 14 = 2(6) + 2
resulting in hexane becoming saturated.
Cyclohexane(C6H12)
C = 6 and H = 12 do not equal 14 (x)!
cyclohexane is an unsaturated molecule as a result.
Cycloalkanes have the general formula C2H2n as well.
Hence, the given statement is false.
Learn more about the hydrocarbons here,
brainly.com/question/17578846
# SPJ4
Answer:
Following are the responses to the given choices:
Explanation:
- The RBC crenation is implied through NaCl by 2,67 percent(m/v) because that solution becomes hypertonic to RBC because of the water within the RBC that passes externally towards the outskirts. RBC thus shrinks.
- 1.13% (m/v), because the low concentration or osmotic that all this solution shows is hypotonic regarding RBC because of the water which has reached the resulting swelling in RBC.
- Distilled H2 implies hemolytic distillation.
- Glucose is indicated by crenation at 8.69 percent (m/v).
- 5.0% (m/v) glucose and 0.9% (m/v) (Crenation is indicated by NaCl.v)
Answer:
3.82 x 10²¹ molecules As₂O₃
Explanation:
To find the amount of molecules arsenic (III) oxide (As₂O₃), you need to (1) convert kg to lbs, then (2) convert g As₂O₃ to moles As₂O₃ (via molar mass), and then (3) convert moles to molecules (via Avogadro's number).
1 kilogram = 2.2 lb
Molar Mass (As₂O₃): 2(74.992 g/mol) + 3(15.998 g/mol)
Molar Mass (As₂O₃): 197.978 g/mol
Avogadro's Number:
6.022 x 10²³ molecules = 1 mole
0.0146 g As₂O₃ 1 kg 189 lb
------------------------ x --------------- x ------------------ x ................
1 kg 2.2 lb
1 mole 6.022 x 10²³ molecules
x ------------------ x --------------------------------------- = 3.82 x 10²¹ molecules As₂O₃
197.978 g 1 mole
