Question
Download Solution PDFRefrigerant R–717 is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
|
|
Saturated Hydrocarbons |
Un-saturated Hydrocarbons |
|
Representation: |
R - (m - 1) (n + 1) P |
R - 1(m - 1) (n + 1) P |
|
Chemical formula: |
CmHnFpClq |
CmHnFpClq |
|
Relation: |
n + p + q = 2m + 2 |
n + p + q = 2m |
where m = No. of Carbon atoms, n = No. of Hydrogen atoms, p = No. of Florine atoms, q = No. of Chlorine atoms.
If the refrigerant is Inorganic Compound
R – (Molecular Weight + 700)
Ammonia NH3 whose molecular weight is 17 and the chemical formula is R – 717.
Water H20 whose molecular weight is 18 and the chemical formula is R – 718.
Carbon dioxide CO2 whose molecular weight is 44 and the chemical formula is R – 744.
Last updated on Jun 23, 2025
-> JKSSB JE application form correction facility has been started. Candidates can make corrections in the JKSSB recruitment 2025 form from June 23 to 27.
-> JKSSB Junior Engineer recruitment exam date 2025 for Civil and Electrical Engineering has been released on its official website.
-> JKSSB JE exam will be conducted on 10th August (Civil), and on 27th July 2025 (Electrical).
-> JKSSB JE recruitment 2025 notification has been released for Civil Engineering.
-> A total of 508 vacancies has been announced for JKSSB JE Civil Engineering recruitment 2025.
-> JKSSB JE Online Application form will be activated from 18th May 2025 to 16th June 2025
-> Candidates who are preparing for the exam can access the JKSSB JE syllabus PDF from official website of JKSSB.
-> The candidates can check the JKSSB JE Previous Year Papers to understand the difficulty level of the exam.
-> Candidates also attempt the JKSSB JE Mock Test which gives you an experience of the actual exam.