SSC Chemistry 14th week Assignment Answer 2022

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SSC Chemistry 14th week Assignment Answer 2022

SSC Chemistry Assignment 2022 has been published. We Prepared all week (14th,13th,12th, 11th, 10th, 9th, 8th, 7th, 6th, 5th, 4th, 3rd, and 1st week) assignment Answer has been published for SSC Chemistry Assignment. Two assignments have been selected from the second chapter. The title of the second chapter is Cells and Tissues. So the solution or answer of 2 assignments from living cells and tissues has to be given. We will provide you with instructions on how to do all week SSC Chemistry assignments. We will also create an assignment and provide a sample for students who do not understand how to do the SSC assignment. You can download all week assignments of SSC Chemistry from our website.

SSC Chemistry 7th Week Assignment 2022

SSC 2022 Chemistry Assignment is scheduled for Science Group students. Students have to prepare SSC Chemistry Assignment Answer 2022 for a total of 14th weeks. The 1st-week assignment has already been published. Chemistry subject has been assigned for the 1st week. It has also been selected for the 3rd, 4th, 6th, 7th, 9th, 10th, and 14th weeks. The 2022 SSC examinee will have to prepare a total of 8 assignment solutions for Chemistry subjects and submit it to the school.
SSC Chymistry 14th Week Assignment 2022

SSC Chemistry Assignment Answer 2022

SSC Chemistry Assignment 2022 Answer, If required, they can be taken help from their teachers, the internet, and others resources. Here you will find the SSC 2022 Chemistry Assignment Answer for all weeks. We will publish Chemistry Answer randomly for all weeks. You will also find next week's assignment solution here. You can create your Chemistry assignment with ideas from our solution.
āωāϤ্āϤāϰ āϏāĻŽূāĻš
āϤাāϰিāĻ–: ā§Ļā§Š āĻŽাāϰ্āϚ, ⧍ā§Ļ⧍⧍
āĻŦāϰাāĻŦāϰ,
āĻĒ্āϰāϧাāύ āĻļিāĻ•্āώāĻ•,
XYZ।
āύāĻŦাāĻŦāĻ—āĻž্āϜ, āĻĻিāύাāϜāĻĒুāϰ।

āĻŦিāώāϝ়: “āĻĒāϰ্āϝাāϝ় āϏাāϰāĻŖিāϰ āĻŽৌāϞেāϰ āĻĒāϰ্āϝাāϝ়āĻŦৃāϤ্āϤিāĻ• āϧāϰ্āĻŽ” āϏāĻŽ্āĻĒāϰ্āĻ•িāϤ| āĻĒ্āϰāϤিāĻŦেāĻĻāύ।
 
āϜāύাāĻŦ, āĻŦিāύীāϤ āύিāĻŦেāĻĻāύ āĻāχ āϝে, āφāĻĒāύাā§ą āφāĻĻেāĻļ āύং āϏ.āχ.āω.āĻŦি ā§Šā§§ā§¯/ā§Ŧ āϤাāϰিāĻ–: ā§Ļā§Š āĻŽাāϰ্āϚ, ⧍ā§Ļ⧍⧍ āĻ…āύুāϏাāϰে “āĻĒāϰ্āϝাāϝ় āϏাāϰāĻŖিāϰ āĻŽৌāϞেāϰ āĻĒāϰ্āϝাāϝ়āĻŦৃāϤ্āϤিāĻ• āϧāϰ্āĻŽ ” āϏāĻŽ্āĻĒāϰ্āĻ•িāϤ āĻĒ্āϰāϤিāĻŦেāĻĻāύāϟি āύিāϝ়ে āĻĒেāĻļ āĻ•āϰāĻ›ি।
āĻĒāϰ্āϝাāϝ āĻ“ āĻ—্āϰুāĻĒেāϰ āϧাāϤāĻŦ āϧāϰ্āĻŽ āύিāϰ্āĻŖāϝ়ঃ
āωāĻ•্āϤ āĻ–āύ্āĻĄিāϤ āϏাāϰāĻŖিāϰ āĻ—্āϰুāĻĒ-ā§§ āĻ“ ā§Š āύāĻŽ্āĻŦāϰ āĻĒāϰ্āϝাāϝ়েāϰ। āĻŽৌāϞāĻ—ুāϞােāϰ āϧাāϤāĻŦ āϧāϰ্āĻŽ āύিāϰ্āĻŖāϝ় āĻ•āϰা āĻšāϞােঃ
āĻ—্āϰুāĻĒ-ā§§ āĻāϰ āĻŽৌāϞ āĻ—ুāϞাে āĻšāϞােঃ Li, Na, K, Rb, Cs āĻāϰ āϧাāϤāĻŦ āϧāϰ্āĻŽ āĻĒāϰ্āϝাāϝ়āĻ•্āϰāĻŽে āφāĻŦāϰ্āϤিāϤ āĻšāϝ়। āφāĻŽāϰা āϜাāύি, āĻ—্āϰুāĻĒ āĻ…āύুāϝাāϝ়ী āύিāϚেāϰ āĻĻিāĻ•ে āĻ—েāϞে āĻŽৌāϞ āĻ—ুāϞােāϰ āĻĒাāϰāĻŽাāĻŖāĻŦিāĻ• āϏংāĻ–্āϝা āĻŦৃāĻĻ্āϧিāϰ āϏাāĻĨে āϏাāĻĨে āύāϤুāύ āĻļāĻ•্āϤিāϏ্āϤāϰ āϝুāĻ•্āϤ āĻšāϝ় āĻāĻŦং āĻĒাāϰāĻŽাāĻŖāĻŦিāĻ• āφāĻ•াāϰ āĻŦৃāĻĻ্āϧি āĻĒাāϝ়। āĻĢāϞে āĻ•েāύ্āĻĻ্āϰ āĻĨেāĻ•ে āϏāϰ্āĻŦāĻŦāĻšিঃāϏ্āĻĨ āϏ্āϤāϰেāϰ āχāϞেāĻ•āϟ্āϰāύেāϰ āĻĻূāϰāϤ্āĻŦ āĻŦৃāĻĻ্āϧি āĻĒাāϝ়। āϝাāϰ āĻ•াāϰāĻŖে āĻŽৌāϞ āĻ—ুāϞাে āϤাāĻĻেāϰ āχāϞেāĻ•āϟ্āϰāύ āĻ–ুāĻŦ āϏāĻšāϜেāχ āϤ্āϝাāĻ— āĻ•āϰāϤে āĻĒাāϰে। āϤাāχ āĻ—্āϰুāĻĒ āĻ…āύুāϝাāϝ়ী
āϝāϤ āύিāϚে āύাāĻŽা āϝাāϝ় āϤāϤāχ āĻŽৌāϞāĻ—ুāϞােāϰ āϧাāϤāĻŦ āϧāϰ্āĻŽ āĻŦাāĻĄ়āϤে āĻĨাāĻ•ে।
āϧাāϤāĻŦ āϧāϰ্āĻŽেāϰ āĻ•্āϰāĻŽ āĻšāϞােঃ Li<Na<K<Rb<Csā§Šāϝ় āĻĒāϰ্āϝাāϝেāϰ āĻŽৌāϞ āĻšāϞােঃ Na, Mg, P, cl āĻāϰা āĻāĻ•āχ āĻĒāϰ্āϝাāϝ়েāϰ āĻŽৌāϞ। āĻāĻĻেāϰ āχāϞেāĻ•āϟ্āϰāύ āĻŦৃāĻĻ্āϧিāϰ āϏাāĻĨে āϏাāĻĨে āύāϤুāύ āĻ•োāύ āĻļāĻ•্āϤিāϏ্āϤāϰ āϝুāĻ•্āϤ āĻšāϝ়āύা। āϤাāχ āĻ•েāύ্āĻĻ্āϰেāϰ āφāĻ•āϰ্āώāĻŖ āϧীāϰে āϧীāϰে āĻŦৃāĻĻ্āϧি āĻĒাāϝ় āĻāĻŦং āĻĒাāϰāĻŽাāĻŖāĻŦিāĻ• āφāĻ•াāϰ āĻš্āϰাāϏ āĻĒাāϝ়। āϤাāχ āϤাāĻĻেāϰ āχāϞেāĻ•āϟ্āϰāύ āϤ্āϝাāĻ— āĻ•āϰাāϰ āĻ•্āώāĻŽāϤা āĻš্āϰাāϏ āĻĒাāϝ়। āĻŽৌāϞāĻ—ুāϞােāϰ āϧাāϤāĻŦ āϧāϰ্āĻŽেāϰ āĻ•্āϰāĻŽ āĻšāϞােঃ Na>Mg>P>CI
āĻĒāϰ্āϝাāϝ় āĻ“ āĻ—্āϰুāĻĒেāϰ āφāϝুāύিāĻ•āϰāĻŖ āĻļāĻ•্āϤি āύিāϰ্āĻŖāϝ়ঃ
āĻ—্āϝাāϏীāϝ় āĻ…āĻŦāϏ্āĻĨাāϝ় āĻāĻ• āĻŽােāϞ āϝাāϝ় āĻĒāϰāĻŽাāĻŖু āĻĨেāĻ•ে āĻāĻ• āĻŽােāϞ āχāϞেāĻ•āϟ্āϰāύ āĻ…āĻĒāϏাāϰāĻŖ āĻ•āϰে āĻāĻ• āĻŽােāϞ āϧāύাāϤ্āĻŽāĻ• āφāύে āĻĒāϰিāĻŖāϤ āĻ•āϰāϤে āϝে āĻĒāϰিāĻŽাāĻŖ āĻļāĻ•্āϤিāϰ āĻĒ্āϰāϝােāϜāύ āĻšāϝ়, āϤাāĻ•ে āϐ āĻŽৌāϞেāϰ āφāϝ়āύিāĻ•āϰāĻŖ āĻļāĻ•্āϤি āĻŦāϞে। |
āωāĻ•্āϤ āĻ–āύ্āĻĄিāϤ āϏাāϰāĻŖিāϰ āĻ—্āϰুāĻĒ-⧍ āĻ“ ā§Ēāϰ্āĻĨ āύāĻŽ্āĻŦāϰ āĻĒāϰ্āϝাāϝ়েāϰ āĻŽৌāϞāĻ—ুāϞােāϰ āφāϝ়āύিāĻ•āϰāĻŖ āĻļāĻ•্āϤি āύিāϰ্āĻŖāϝ় āĻ•āϰা āĻšāϞােঃ
āĻ—্āϰুāĻĒ-⧍ āĻāϰ āĻŽৌāϞ āĻ—ুāϞাে āĻšāϞাে: Be, Mg, ca, Sr, Ba āĻāϰ āφāϝ়āύিāĻ•āϰāĻŖ āĻļāĻ•্āϤি āĻĒāϰ্āϝাāϝ়āĻ•্āϰāĻŽে āφāĻŦāϰ্āϤিāϤ āĻšāϝ়। āφāĻŽāϰা āϜাāύি, āĻ—্āϰুāĻĒ āĻ…āύু্āϝাāϝ়ী āύিāϚেāϰ āĻĻিāĻ•ে āĻ—েāϞে āĻŽৌāϞ āĻ—ুāϞােāϰ āĻĒাāϰāĻŽাāĻŖāĻŦিāĻ• āϏংāĻ–্āϝা āĻŦৃāĻĻ্āϧিāϰ āϏাāĻĨে āϏাāĻĨে āĻĒাāϰāĻŽাāĻŖāĻŦিāĻ• āφāĻ•াāϰ āĻŦৃāĻĻ্āϧি āĻĒাāϝ়। āϝাāϰ āĻ•াāϰāĻŖে āĻŽৌāϞ āĻ—ুāϞাে āϤাāĻĻেāϰ āχāϞেāĻ•āϟ্āϰāύ āĻ–ুāĻŦ āϏāĻšāϜেāχ āϤ্āϝাāĻ— āĻ•āϰāϤে āĻĒাāϰে। āφāĻŦাāϰ āĻŽৌāϞেāϰ āφāĻ•াāϰ āĻŦৃāĻĻ্āϧি āĻĒেāϞে āĻŽৌāϞেāϰ āφāϝ়āύিāĻ•āϰāĻŖ āĻļāĻ•্āϤি āĻ•āĻŽে। āĻ­াāχ āĻ—্āϰুāĻĒ āĻ…āύু্āϝাāϝ়ী āϝāϤ āύিāϚে āύাāĻŽা āϝাāϝ় āϤāϤāχ āĻŽৌāϞāĻ—ুāϞােāϰ āφāϝ়āύিāĻ•āϰāĻŖ āĻļāĻ•্āϤি āĻ•āĻŽāϤে āĻĨাāĻ•ে।
ā§Ēāϰ্āĻĨ āĻĒāϰ্āϝাāϝেāϰ āĻŽৌāϞ āĻšāϞােঃ K, Ca, As, Br āĻāϰা āĻāĻ•āχ āĻĒāϰ্āϝাāϝেāϰ āĻŽৌāϞ। āĻāĻĻেāϰ āχāϞেāĻ•āϟ্āϰāύ āĻŦৃāĻĻ্āϧিāϰ āϏাāĻĨে āϏাāĻĨে āύāϤুāύ āĻ•োāύ āĻļāĻ•্āϤিāϏ্āϤāϰ āϝুāĻ•্āϤ āĻšāϝ়āύা। āϤাāχ āĻ•েāύ্āĻĻ্āϰেāϰ āφāĻ•āϰ্āώāĻŖ āϧীāϰে āϧীāϰে āĻŦৃāĻĻ্āϧি āĻĒাāϝ় āĻāĻŦং āĻĒাāϰāĻŽাāĻŖāĻŦিāĻ• āφāĻ•াāϰ āĻš্āϰাāϏ āĻĒাāϝ়। āϤাāχ āϤাāĻĻেāϰ āϤāĻĄ়িā§Ž āĻ‹āĻŖাāϤ্āĻŽāĻ•āϤা āĻŦৃāĻĻ্āϧি āĻĒাāϝ়। āϤāĻĄ়িā§Ž āĻ‹āĻŖাāϤ্āĻŽāĻ•āϤাāϰ āĻ•্āϰāĻŽ āĻšāϞােঃ K<Ca<As<Br

āĻ•্āώাāϰāϧাāϤুāϏāĻŽূāĻšেāϰ āϝৌāĻ—েāϰ āĻāĻ•āχ āϰāĻ•āĻŽ āĻŦিāĻ•্āϰিāϝ়া āĻĒ্āϰāĻĻāϰ্āĻļāύঃ
āĻ—্āϰুāĻĒ-ā§§ āĻāϰ āĻŽৌāϞ āĻ—ুāϞােāĻ•ে āĻ•্āώাāϰ āϧাāϤু āĻŦāϞা āĻš্āϝ। āĻ•াāϰāĻŖ | āĻāϰা āĻĒাāύিāϰ āϏাāĻĨে āĻŦিāĻ•্āϰিāϝ়া āĻ•āϰে āĻ•্āώাāϰ āĻ“ H, āĻ—্āϝাāϏ āϤৈāϰি āĻ•āϰে। āĻ—্āϰুāĻĒ-ā§§ āĻāϰ āĻ•্āώাāϰ āϧাāϤুāĻ—ুāϞাে āĻšāϞােঃ Na, K, Rb, cs, Fr. | āĻāĻĻেāϰ āϏāϰ্āĻŦāĻļেāώ āĻ•āĻ•্āώāĻĒāĻĨে 1āϟি āĻ•āϰে āχāϞেāĻ•āϟ্āϰāύ āĻĨাāĻ•ে। āϤাāχ āϤাāϰা āϏāĻšāϜে 1āϟি āχāϞেāĻ•āϟ্āϰāύ āϤ্āϝাāĻ— āĻ•āϰে āφāϝ়āύিāĻ• āϝৌāĻ— āϤৈāϰি āĻ•āϰāϤে āĻĒাāϰে। āĻāϰা āĻĒাāύিāϰ āϏাāĻĨে āĻŦিāĻ•্āϰি āĻ•āϰে āĻ•্āώাāϰ āĻ“ āĻšাāχāĻĄ্āϰোāϜেāύ āĻ—্āϝাāϏ āĻ‰ā§ŽāĻĒāύ্āύ āĻ•āϰে। 2Li + 2Hā§Ļ = 2LiOH + H, | 2Na + 2HO = 2NaOH + H2 2K + 2Hā§Ļ = 2KOH + H, 2Rb + 2HO = 2RbOH + H, 
āφāĻŦাāϰ āĻāĻĻেāϰ āĻ•্āώাāϰ āĻāϏিāĻĄেāϰ āϏাāĻĨে āĻŦিāĻ•্āϰি āĻ•āϰে āϞāĻŦāĻŖ āĻ“ āĻĒাāύি āĻ‰ā§ŽāĻĒāύ্āύ āĻ•āϰে।
LiOH + HCl = LiCl + HO NaOH + HCI = NaCl + H2O KOH + HCl = KCI + HO bOH + HCl = RbCl + Hā§Ļ āĻāĻĻেāϰ āϧāϰ্āĻŽ āĻāĻ•āχ āĻšāĻ“āϝ়াāϝ় āĻāϰা āϏāϰ্āĻŦāĻĻা āĻāĻ•āχ āĻŦিāĻ•্āϰিāϝ়া āĻĒ্āϰāĻĻāϰ্āĻļāύ āĻ•āϰে। āĻ•্āώাāϰ āϧাāϤুāϏāĻŽূāĻš āĻāĻ•āχ āĻŦিāĻ•্āϰিāϝ়া āĻĒ্āϰāĻĻāϰ্āĻļāύ āĻ•āϰে।

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