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.
āĻ‰āĻ¤্āĻ¤āĻ° āĻ¸āĻŽূāĻš
āĻ¤াāĻ°িāĻ–: ā§Ļā§Š āĻŽাāĻ°্āĻš, ā§¨ā§Ļā§¨ā§¨
āĻŦāĻ°াāĻŦāĻ°,
āĻĒ্āĻ°āĻ§াāĻ¨ āĻļিāĻ•্āĻˇāĻ•,
XYZ।
āĻ¨āĻŦাāĻŦāĻ—āĻž্āĻœ, āĻĻিāĻ¨াāĻœāĻĒুāĻ°।

āĻŦিāĻˇāĻ¯়: “āĻĒāĻ°্āĻ¯াāĻ¯় āĻ¸াāĻ°āĻŖিāĻ° āĻŽৌāĻ˛েāĻ° āĻĒāĻ°্āĻ¯াāĻ¯়āĻŦৃāĻ¤্āĻ¤িāĻ• āĻ§āĻ°্āĻŽ” āĻ¸āĻŽ্āĻĒāĻ°্āĻ•িāĻ¤| āĻĒ্āĻ°āĻ¤িāĻŦেāĻĻāĻ¨।
 
āĻœāĻ¨াāĻŦ, āĻŦিāĻ¨ীāĻ¤ āĻ¨িāĻŦেāĻĻāĻ¨ āĻāĻ‡ āĻ¯ে, āĻ†āĻĒāĻ¨াā§ą āĻ†āĻĻেāĻļ āĻ¨ং āĻ¸.āĻ‡.āĻ‰.āĻŦি ā§Šā§§ā§¯/ā§Ŧ āĻ¤াāĻ°িāĻ–: ā§Ļā§Š āĻŽাāĻ°্āĻš, ā§¨ā§Ļā§¨ā§¨ āĻ…āĻ¨ুāĻ¸াāĻ°ে “āĻĒāĻ°্āĻ¯াāĻ¯় āĻ¸াāĻ°āĻŖিāĻ° āĻŽৌāĻ˛েāĻ° āĻĒāĻ°্āĻ¯াāĻ¯়āĻŦৃāĻ¤্āĻ¤িāĻ• āĻ§āĻ°্āĻŽ ” āĻ¸āĻŽ্āĻĒāĻ°্āĻ•িāĻ¤ āĻĒ্āĻ°āĻ¤িāĻŦেāĻĻāĻ¨āĻŸি āĻ¨িāĻ¯়ে āĻĒেāĻļ āĻ•āĻ°āĻ›ি।
āĻĒāĻ°্āĻ¯াāĻ¯ āĻ“ āĻ—্āĻ°ুāĻĒেāĻ° āĻ§াāĻ¤āĻŦ āĻ§āĻ°্āĻŽ āĻ¨িāĻ°্āĻŖāĻ¯়ঃ
āĻ‰āĻ•্āĻ¤ āĻ–āĻ¨্āĻĄিāĻ¤ āĻ¸াāĻ°āĻŖিāĻ° āĻ—্āĻ°ুāĻĒ-ā§§ āĻ“ ā§Š āĻ¨āĻŽ্āĻŦāĻ° āĻĒāĻ°্āĻ¯াāĻ¯়েāĻ°। āĻŽৌāĻ˛āĻ—ুāĻ˛ােāĻ° āĻ§াāĻ¤āĻŦ āĻ§āĻ°্āĻŽ āĻ¨িāĻ°্āĻŖāĻ¯় āĻ•āĻ°া āĻšāĻ˛ােঃ
āĻ—্āĻ°ুāĻĒ-ā§§ āĻāĻ° āĻŽৌāĻ˛ āĻ—ুāĻ˛াে āĻšāĻ˛ােঃ Li, Na, K, Rb, Cs āĻāĻ° āĻ§াāĻ¤āĻŦ āĻ§āĻ°্āĻŽ āĻĒāĻ°্āĻ¯াāĻ¯়āĻ•্āĻ°āĻŽে āĻ†āĻŦāĻ°্āĻ¤িāĻ¤ āĻšāĻ¯়। āĻ†āĻŽāĻ°া āĻœাāĻ¨ি, āĻ—্āĻ°ুāĻĒ āĻ…āĻ¨ুāĻ¯াāĻ¯়ী āĻ¨িāĻšেāĻ° āĻĻিāĻ•ে āĻ—েāĻ˛ে āĻŽৌāĻ˛ āĻ—ুāĻ˛ােāĻ° āĻĒাāĻ°āĻŽাāĻŖāĻŦিāĻ• āĻ¸ংāĻ–্āĻ¯া āĻŦৃāĻĻ্āĻ§িāĻ° āĻ¸াāĻĨে āĻ¸াāĻĨে āĻ¨āĻ¤ুāĻ¨ āĻļāĻ•্āĻ¤িāĻ¸্āĻ¤āĻ° āĻ¯ুāĻ•্āĻ¤ āĻšāĻ¯় āĻāĻŦং āĻĒাāĻ°āĻŽাāĻŖāĻŦিāĻ• āĻ†āĻ•াāĻ° āĻŦৃāĻĻ্āĻ§ি āĻĒাāĻ¯়। āĻĢāĻ˛ে āĻ•েāĻ¨্āĻĻ্āĻ° āĻĨেāĻ•ে āĻ¸āĻ°্āĻŦāĻŦāĻšিঃāĻ¸্āĻĨ āĻ¸্āĻ¤āĻ°েāĻ° āĻ‡āĻ˛েāĻ•āĻŸ্āĻ°āĻ¨েāĻ° āĻĻূāĻ°āĻ¤্āĻŦ āĻŦৃāĻĻ্āĻ§ি āĻĒাāĻ¯়। āĻ¯াāĻ° āĻ•াāĻ°āĻŖে āĻŽৌāĻ˛ āĻ—ুāĻ˛াে āĻ¤াāĻĻেāĻ° āĻ‡āĻ˛েāĻ•āĻŸ্āĻ°āĻ¨ āĻ–ুāĻŦ āĻ¸āĻšāĻœেāĻ‡ āĻ¤্āĻ¯াāĻ— āĻ•āĻ°āĻ¤ে āĻĒাāĻ°ে। āĻ¤াāĻ‡ āĻ—্āĻ°ুāĻĒ āĻ…āĻ¨ুāĻ¯াāĻ¯়ী
āĻ¯āĻ¤ āĻ¨িāĻšে āĻ¨াāĻŽা āĻ¯াāĻ¯় āĻ¤āĻ¤āĻ‡ āĻŽৌāĻ˛āĻ—ুāĻ˛ােāĻ° āĻ§াāĻ¤āĻŦ āĻ§āĻ°্āĻŽ āĻŦাāĻĄ়āĻ¤ে āĻĨাāĻ•ে।
āĻ§াāĻ¤āĻŦ āĻ§āĻ°্āĻŽেāĻ° āĻ•্āĻ°āĻŽ āĻšāĻ˛ােঃ Li<Na<K<Rb<Csā§ŠāĻ¯় āĻĒāĻ°্āĻ¯াāĻ¯েāĻ° āĻŽৌāĻ˛ āĻšāĻ˛ােঃ Na, Mg, P, cl āĻāĻ°া āĻāĻ•āĻ‡ āĻĒāĻ°্āĻ¯াāĻ¯়েāĻ° āĻŽৌāĻ˛। āĻāĻĻেāĻ° āĻ‡āĻ˛েāĻ•āĻŸ্āĻ°āĻ¨ āĻŦৃāĻĻ্āĻ§িāĻ° āĻ¸াāĻĨে āĻ¸াāĻĨে āĻ¨āĻ¤ুāĻ¨ āĻ•োāĻ¨ āĻļāĻ•্āĻ¤িāĻ¸্āĻ¤āĻ° āĻ¯ুāĻ•্āĻ¤ āĻšāĻ¯়āĻ¨া। āĻ¤াāĻ‡ āĻ•েāĻ¨্āĻĻ্āĻ°েāĻ° āĻ†āĻ•āĻ°্āĻˇāĻŖ āĻ§ীāĻ°ে āĻ§ীāĻ°ে āĻŦৃāĻĻ্āĻ§ি āĻĒাāĻ¯় āĻāĻŦং āĻĒাāĻ°āĻŽাāĻŖāĻŦিāĻ• āĻ†āĻ•াāĻ° āĻš্āĻ°াāĻ¸ āĻĒাāĻ¯়। āĻ¤াāĻ‡ āĻ¤াāĻĻেāĻ° āĻ‡āĻ˛েāĻ•āĻŸ্āĻ°āĻ¨ āĻ¤্āĻ¯াāĻ— āĻ•āĻ°াāĻ° āĻ•্āĻˇāĻŽāĻ¤া āĻš্āĻ°াāĻ¸ āĻĒাāĻ¯়। āĻŽৌāĻ˛āĻ—ুāĻ˛ােāĻ° āĻ§াāĻ¤āĻŦ āĻ§āĻ°্āĻŽেāĻ° āĻ•্āĻ°āĻŽ āĻšāĻ˛ােঃ Na>Mg>P>CI
āĻĒāĻ°্āĻ¯াāĻ¯় āĻ“ āĻ—্āĻ°ুāĻĒেāĻ° āĻ†āĻ¯ুāĻ¨িāĻ•āĻ°āĻŖ āĻļāĻ•্āĻ¤ি āĻ¨িāĻ°্āĻŖāĻ¯়ঃ
āĻ—্āĻ¯াāĻ¸ীāĻ¯় āĻ…āĻŦāĻ¸্āĻĨাāĻ¯় āĻāĻ• āĻŽােāĻ˛ āĻ¯াāĻ¯় āĻĒāĻ°āĻŽাāĻŖু āĻĨেāĻ•ে āĻāĻ• āĻŽােāĻ˛ āĻ‡āĻ˛েāĻ•āĻŸ্āĻ°āĻ¨ āĻ…āĻĒāĻ¸াāĻ°āĻŖ āĻ•āĻ°ে āĻāĻ• āĻŽােāĻ˛ āĻ§āĻ¨াāĻ¤্āĻŽāĻ• āĻ†āĻ¨ে āĻĒāĻ°িāĻŖāĻ¤ āĻ•āĻ°āĻ¤ে āĻ¯ে āĻĒāĻ°িāĻŽাāĻŖ āĻļāĻ•্āĻ¤িāĻ° āĻĒ্āĻ°āĻ¯ােāĻœāĻ¨ āĻšāĻ¯়, āĻ¤াāĻ•ে āĻ āĻŽৌāĻ˛েāĻ° āĻ†āĻ¯়āĻ¨িāĻ•āĻ°āĻŖ āĻļāĻ•্āĻ¤ি āĻŦāĻ˛ে। |
āĻ‰āĻ•্āĻ¤ āĻ–āĻ¨্āĻĄিāĻ¤ āĻ¸াāĻ°āĻŖিāĻ° āĻ—্āĻ°ুāĻĒ-ā§¨ āĻ“ ā§ĒāĻ°্āĻĨ āĻ¨āĻŽ্āĻŦāĻ° āĻĒāĻ°্āĻ¯াāĻ¯়েāĻ° āĻŽৌāĻ˛āĻ—ুāĻ˛ােāĻ° āĻ†āĻ¯়āĻ¨িāĻ•āĻ°āĻŖ āĻļāĻ•্āĻ¤ি āĻ¨িāĻ°্āĻŖāĻ¯় āĻ•āĻ°া āĻšāĻ˛ােঃ
āĻ—্āĻ°ুāĻĒ-ā§¨ āĻāĻ° āĻŽৌāĻ˛ āĻ—ুāĻ˛াে āĻšāĻ˛াে: Be, Mg, ca, Sr, Ba āĻāĻ° āĻ†āĻ¯়āĻ¨িāĻ•āĻ°āĻŖ āĻļāĻ•্āĻ¤ি āĻĒāĻ°্āĻ¯াāĻ¯়āĻ•্āĻ°āĻŽে āĻ†āĻŦāĻ°্āĻ¤িāĻ¤ āĻšāĻ¯়। āĻ†āĻŽāĻ°া āĻœাāĻ¨ি, āĻ—্āĻ°ুāĻĒ āĻ…āĻ¨ু্āĻ¯াāĻ¯়ী āĻ¨িāĻšেāĻ° āĻĻিāĻ•ে āĻ—েāĻ˛ে āĻŽৌāĻ˛ āĻ—ুāĻ˛ােāĻ° āĻĒাāĻ°āĻŽাāĻŖāĻŦিāĻ• āĻ¸ংāĻ–্āĻ¯া āĻŦৃāĻĻ্āĻ§িāĻ° āĻ¸াāĻĨে āĻ¸াāĻĨে āĻĒাāĻ°āĻŽাāĻŖāĻŦিāĻ• āĻ†āĻ•াāĻ° āĻŦৃāĻĻ্āĻ§ি āĻĒাāĻ¯়। āĻ¯াāĻ° āĻ•াāĻ°āĻŖে āĻŽৌāĻ˛ āĻ—ুāĻ˛াে āĻ¤াāĻĻেāĻ° āĻ‡āĻ˛েāĻ•āĻŸ্āĻ°āĻ¨ āĻ–ুāĻŦ āĻ¸āĻšāĻœেāĻ‡ āĻ¤্āĻ¯াāĻ— āĻ•āĻ°āĻ¤ে āĻĒাāĻ°ে। āĻ†āĻŦাāĻ° āĻŽৌāĻ˛েāĻ° āĻ†āĻ•াāĻ° āĻŦৃāĻĻ্āĻ§ি āĻĒেāĻ˛ে āĻŽৌāĻ˛েāĻ° āĻ†āĻ¯়āĻ¨িāĻ•āĻ°āĻŖ āĻļāĻ•্āĻ¤ি āĻ•āĻŽে। āĻ­াāĻ‡ āĻ—্āĻ°ুāĻĒ āĻ…āĻ¨ু্āĻ¯াāĻ¯়ী āĻ¯āĻ¤ āĻ¨িāĻšে āĻ¨াāĻŽা āĻ¯াāĻ¯় āĻ¤āĻ¤āĻ‡ āĻŽৌāĻ˛āĻ—ুāĻ˛ােāĻ° āĻ†āĻ¯়āĻ¨িāĻ•āĻ°āĻŖ āĻļāĻ•্āĻ¤ি āĻ•āĻŽāĻ¤ে āĻĨাāĻ•ে।
ā§ĒāĻ°্āĻĨ āĻĒāĻ°্āĻ¯াāĻ¯েāĻ° āĻŽৌāĻ˛ āĻšāĻ˛ােঃ K, Ca, As, Br āĻāĻ°া āĻāĻ•āĻ‡ āĻĒāĻ°্āĻ¯াāĻ¯েāĻ° āĻŽৌāĻ˛। āĻāĻĻেāĻ° āĻ‡āĻ˛েāĻ•āĻŸ্āĻ°āĻ¨ āĻŦৃāĻĻ্āĻ§িāĻ° āĻ¸াāĻĨে āĻ¸াāĻĨে āĻ¨āĻ¤ুāĻ¨ āĻ•োāĻ¨ āĻļāĻ•্āĻ¤িāĻ¸্āĻ¤āĻ° āĻ¯ুāĻ•্āĻ¤ āĻšāĻ¯়āĻ¨া। āĻ¤াāĻ‡ āĻ•েāĻ¨্āĻĻ্āĻ°েāĻ° āĻ†āĻ•āĻ°্āĻˇāĻŖ āĻ§ীāĻ°ে āĻ§ীāĻ°ে āĻŦৃāĻĻ্āĻ§ি āĻĒাāĻ¯় āĻāĻŦং āĻĒাāĻ°āĻŽাāĻŖāĻŦিāĻ• āĻ†āĻ•াāĻ° āĻš্āĻ°াāĻ¸ āĻĒাāĻ¯়। āĻ¤াāĻ‡ āĻ¤াāĻĻেāĻ° āĻ¤āĻĄ়িā§Ž āĻ‹āĻŖাāĻ¤্āĻŽāĻ•āĻ¤া āĻŦৃāĻĻ্āĻ§ি āĻĒাāĻ¯়। āĻ¤āĻĄ়িā§Ž āĻ‹āĻŖাāĻ¤্āĻŽāĻ•āĻ¤াāĻ° āĻ•্āĻ°āĻŽ āĻšāĻ˛ােঃ K<Ca<As<Br

āĻ•্āĻˇাāĻ°āĻ§াāĻ¤ুāĻ¸āĻŽূāĻšেāĻ° āĻ¯ৌāĻ—েāĻ° āĻāĻ•āĻ‡ āĻ°āĻ•āĻŽ āĻŦিāĻ•্āĻ°িāĻ¯়া āĻĒ্āĻ°āĻĻāĻ°্āĻļāĻ¨ঃ
āĻ—্āĻ°ুāĻĒ-ā§§ āĻāĻ° āĻŽৌāĻ˛ āĻ—ুāĻ˛ােāĻ•ে āĻ•্āĻˇাāĻ° āĻ§াāĻ¤ু āĻŦāĻ˛া āĻš্āĻ¯। āĻ•াāĻ°āĻŖ | āĻāĻ°া āĻĒাāĻ¨িāĻ° āĻ¸াāĻĨে āĻŦিāĻ•্āĻ°িāĻ¯়া āĻ•āĻ°ে āĻ•্āĻˇাāĻ° āĻ“ H, āĻ—্āĻ¯াāĻ¸ āĻ¤ৈāĻ°ি āĻ•āĻ°ে। āĻ—্āĻ°ুāĻĒ-ā§§ āĻāĻ° āĻ•্āĻˇাāĻ° āĻ§াāĻ¤ুāĻ—ুāĻ˛াে āĻšāĻ˛ােঃ Na, K, Rb, cs, Fr. | āĻāĻĻেāĻ° āĻ¸āĻ°্āĻŦāĻļেāĻˇ āĻ•āĻ•্āĻˇāĻĒāĻĨে 1āĻŸি āĻ•āĻ°ে āĻ‡āĻ˛েāĻ•āĻŸ্āĻ°āĻ¨ āĻĨাāĻ•ে। āĻ¤াāĻ‡ āĻ¤াāĻ°া āĻ¸āĻšāĻœে 1āĻŸি āĻ‡āĻ˛েāĻ•āĻŸ্āĻ°āĻ¨ āĻ¤্āĻ¯াāĻ— āĻ•āĻ°ে āĻ†āĻ¯়āĻ¨িāĻ• āĻ¯ৌāĻ— āĻ¤ৈāĻ°ি āĻ•āĻ°āĻ¤ে āĻĒাāĻ°ে। āĻāĻ°া āĻĒাāĻ¨িāĻ° āĻ¸াāĻĨে āĻŦিāĻ•্āĻ°ি āĻ•āĻ°ে āĻ•্āĻˇাāĻ° āĻ“ āĻšাāĻ‡āĻĄ্āĻ°োāĻœেāĻ¨ āĻ—্āĻ¯াāĻ¸ āĻ‰ā§ŽāĻĒāĻ¨্āĻ¨ āĻ•āĻ°ে। 2Li + 2Hā§Ļ = 2LiOH + H, | 2Na + 2HO = 2NaOH + H2 2K + 2Hā§Ļ = 2KOH + H, 2Rb + 2HO = 2RbOH + H, 
āĻ†āĻŦাāĻ° āĻāĻĻেāĻ° āĻ•্āĻˇাāĻ° āĻāĻ¸িāĻĄেāĻ° āĻ¸াāĻĨে āĻŦিāĻ•্āĻ°ি āĻ•āĻ°ে āĻ˛āĻŦāĻŖ āĻ“ āĻĒাāĻ¨ি āĻ‰ā§ŽāĻĒāĻ¨্āĻ¨ āĻ•āĻ°ে।
LiOH + HCl = LiCl + HO NaOH + HCI = NaCl + H2O KOH + HCl = KCI + HO bOH + HCl = RbCl + Hā§Ļ āĻāĻĻেāĻ° āĻ§āĻ°্āĻŽ āĻāĻ•āĻ‡ āĻšāĻ“āĻ¯়াāĻ¯় āĻāĻ°া āĻ¸āĻ°্āĻŦāĻĻা āĻāĻ•āĻ‡ āĻŦিāĻ•্āĻ°িāĻ¯়া āĻĒ্āĻ°āĻĻāĻ°্āĻļāĻ¨ āĻ•āĻ°ে। āĻ•্āĻˇাāĻ° āĻ§াāĻ¤ুāĻ¸āĻŽূāĻš āĻāĻ•āĻ‡ āĻŦিāĻ•্āĻ°িāĻ¯়া āĻĒ্āĻ°āĻĻāĻ°্āĻļāĻ¨ āĻ•āĻ°ে।

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