{"id":36275,"date":"2024-02-19T09:04:33","date_gmt":"2024-02-19T07:04:33","guid":{"rendered":"https:\/\/www.nsa.smsm.lt\/old\/?page_id=36275"},"modified":"2024-09-24T08:56:47","modified_gmt":"2024-09-24T06:56:47","slug":"chemija","status":"publish","type":"page","link":"https:\/\/www.nsa.smsm.lt\/old\/egzaminai-ir-pasiekimu-patikrinimai\/2024-2025-m-pasiekimu-patikrinimai\/chemija\/","title":{"rendered":"Chemija"},"content":{"rendered":"

2024 – 2025 m. m.<\/h2>\n

Chemijos valstybinio brandos egzamino pirma dalis. U\u017eduoties pavyzdys<\/a><\/p>\n

Chemijos valstybinio brandos egzamino pirmos dalies kandidat\u0173 darb\u0173 vertinimo instrukcija<\/a><\/p>\n

Chemijos valstybinio brandos egzamino antros dalies u\u017eduotis<\/a><\/p>\n

Chemijos valstybinio brandos egzamino antros dalies kandidat\u0173 darb\u0173 vertinimo instrukcija<\/a><\/p>\n

2023 – 2024 m. m.<\/h2>\n

Chemijos tarpinio patikrinimo u\u017eduoties pavyzdys<\/a><\/p>\n

Chemijos tarpinio patikrinimo u\u017eduoties pavyzd\u017eio vertinimo instrukcija<\/a><\/p>\n

Chemijos valstybinio brandos egzamino lenteli\u0173 rinkinys<\/a><\/p>\n

Pastaba: siekiant ai\u0161kumo d\u0117l reik\u0161mini\u0173 skaitmen\u0173 naudojimo sutarta:<\/strong><\/p>\n

1)\u00a0 atom\u0173 santykini\u0173 masi\u0173 reik\u0161mes apvalinti:<\/p>\n

1\u20135 klas\u0117se \u2013\u00a0iki sveiko\u00a0skai\u010diaus,<\/p>\n

6\u20138 klas\u0117se \u2013\u00a0iki de\u0161imt\u0173j\u0173,<\/p>\n

9 (I gimnazijos)\u2013IV gimnazijos\u00a0klas\u0117se \u2013 iki \u0161imt\u0173j\u0173;<\/p>\n

2) 2024\u20132025 m. Matematikos, gamtos moksl\u0173 dalyk\u0173, informatikos ir in\u017einerini\u0173 technologij\u0173 valstybini\u0173 brandos egzamin\u0173 u\u017eduo\u010di\u0173 apra\u0161o 2 priedo periodin\u0117je element\u0173 lentel\u0117je atom\u0173 santykines mases pateikti\u00a0\u0161imt\u0173j\u0173 tikslumu\u00a0\u2013 du skaitmenys po kablelio, pavyzd\u017eiui, vietoj Ar<\/sub>(N) = 14,0067 ra\u0161yti Ar<\/sub>(N) = 14,01.<\/p>\n","protected":false},"excerpt":{"rendered":"

2024 – 2025 m. m. Chemijos valstybinio brandos egzamino pirma dalis. U\u017eduoties pavyzdys Chemijos valstybinio brandos egzamino pirmos dalies kandidat\u0173 darb\u0173 vertinimo instrukcija Chemijos valstybinio brandos egzamino antros dalies u\u017eduotis Chemijos valstybinio brandos egzamino antros dalies kandidat\u0173 darb\u0173 vertinimo instrukcija Pla\u010diau <\/i><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":34932,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"page-2024pasiekimas.php","meta":{"_acf_changed":false,"footnotes":""},"class_list":["post-36275","page","type-page","status-publish","hentry"],"acf":[],"_links":{"self":[{"href":"https:\/\/www.nsa.smsm.lt\/old\/wp-json\/wp\/v2\/pages\/36275","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.nsa.smsm.lt\/old\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.nsa.smsm.lt\/old\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.nsa.smsm.lt\/old\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.nsa.smsm.lt\/old\/wp-json\/wp\/v2\/comments?post=36275"}],"version-history":[{"count":10,"href":"https:\/\/www.nsa.smsm.lt\/old\/wp-json\/wp\/v2\/pages\/36275\/revisions"}],"predecessor-version":[{"id":41031,"href":"https:\/\/www.nsa.smsm.lt\/old\/wp-json\/wp\/v2\/pages\/36275\/revisions\/41031"}],"up":[{"embeddable":true,"href":"https:\/\/www.nsa.smsm.lt\/old\/wp-json\/wp\/v2\/pages\/34932"}],"wp:attachment":[{"href":"https:\/\/www.nsa.smsm.lt\/old\/wp-json\/wp\/v2\/media?parent=36275"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}