A) 1s on H and 3p on Cl
In HCl, the H atom has only one valance electron. Each share an electron an therefore a single covalent bond is formed between the two. The bond in HCl is therefore a result of an overlap between 1s orbital and ONLY ONE of the lobes of the 3p orbital of Chlorine.
We have that the Complete Expanded Structure of (CH3)2CHCH2OCH2CH3 is given in the attachment below
From the Question
(CH3)2CHCH2OCH2CH3
Generally for the condensed formula (CH3)2CHCH2OCH2CH3
We consider that this is a single bond connecting them
We consider
Hydrogen H(1)
Oxygen(8)
Carbon(6)
In conclusion
The Complete Expanded Structure of (CH3)2CHCH2OCH2CH3 is given in the attachment below.
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3I₂ + 2Al → 2AlI₃
m(I₂)=3M(I₂)m(Al)/{2M(Al)}
m(I₂)=3*253.8*20.4/{2*27.0}=287.64 g
Answer:
A. percentage mass of iron = 5.17%
percentage mass of sand = 8.62%
percentage mass of water = 86.205%
B. (Iron + sand + water) -------> ( iron + sand) ------> sand
C. The step of separation of iron and sand
Explanation:
A. Percentage mass of the mixtures:
Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g
percentage mass of iron = 15/290 * 100% = 5.17%
percentage mass of sand = 25/290 * 100% = 8.62%
percentage mass of water = 250/290 * 100% = 86.205%
B. Flow chart of separation procedure
(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand
C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.
Answer:
If the volume is doubled and the number of molecules is doubled, pressure is unchanged
Explanation:
Step 1: Data given
Temperature = constant
Volume will be doubled
Number of molecules will be doubles
Step 2:
p*V = n*R*T
⇒ gas constant and temperature are constant
Initial pressure = n*R*T / V
Initial pressure = 2*R*T/2
Initial pressure = RT
Final pressure = 4*RT / 4
Final pressure = R*T
If the volume is doubled and the number of molecules is doubled, pressure is unchanged
