Answer:
When the pressure increases to 2.35 atm, the temperature will increase to 378 K
Explanation:
Step 1: Data given
The initial pressure = 1.82 atm
The initial temperature = 293 K
The pressure will be increased to 2.35 atm
Step 2: Calculate the new temperature
P1/T1 = P2/T2
⇒with P1 = the initial pressure = 1.82 atm
⇒with T1 = the initial temperature = 293 K
⇒with P2 = the increased pressure = 2.35 atm
⇒with T2 = the new temperature = TO BE DETERMINED
1.82atm / 293 K = 2.35 atm / T2
T2 = 2.35 atm / (1.82 atm/293 K)
T2 = 2.35 / 0.0062116
T2 = 378 K
When the pressure increases to 2.35 atm, the temperature will increase to 378 K
Answer:
Degrees Kelvin = Degrees Celsius + 273.15
Degrees Kelvin = -13 + 273.15
= 260.15
°
Explanation:
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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Answer:
Group 1 (or IA)
Explanation:
If element X is a halogen, then it belongs to the group 17 (or VIIA, under a different notation).
For each extra unit of atomic number, the group number increases by 1. That means that the X+1 element would belong to the group 18 (or VIIIA). <em>The X+2 element would thus belong in the group 1 </em>(or IA) one period higher (higher as in numeric value, not as in position in the periodic table).
