{"id":567428,"date":"2026-07-22T09:22:08","date_gmt":"2026-07-22T02:22:08","guid":{"rendered":"https:\/\/hoanghamobile.com\/tin-tuc\/?p=567428"},"modified":"2026-07-23T09:55:22","modified_gmt":"2026-07-23T02:55:22","slug":"nguyen-tu-khoi","status":"publish","type":"post","link":"https:\/\/hoanghamobile.com\/tin-tuc\/nguyen-tu-khoi\/","title":{"rendered":"Nguy\u00ean t\u1eed kh\u1ed1i l\u00e0 g\u00ec? B\u1ea3ng nguy\u00ean t\u1eed kh\u1ed1i c\u00e1c nguy\u00ean t\u1ed1"},"content":{"rendered":"

Nguy\u00ean t\u1eed kh\u1ed1i l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed9t nguy\u00ean t\u1eed t\u00ednh b\u1eb1ng \u0111\u01a1n v\u1ecb cacbon (\u0111vC), trong \u0111\u00f3 1 \u0111vC b\u1eb1ng 1\/12 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a nguy\u00ean t\u1eed cacbon.<\/strong> Nh\u1edd nguy\u00ean t\u1eed kh\u1ed1i, ta c\u00f3 th\u1ec3 so s\u00e1nh \u0111\u1ed9 n\u1eb7ng nh\u1eb9 gi\u1eefa c\u00e1c nguy\u00ean t\u1eed<\/a> v\u00e0 t\u00ednh ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a c\u00e1c ch\u1ea5t<\/a>. \u0110\u00e2y l\u00e0 ki\u1ebfn th\u1ee9c n\u1ec1n t\u1ea3ng c\u1ee7a H\u00f3a h\u1ecdc l\u1edbp 8. B\u00e0i vi\u1ebft n\u00e0y tr\u00ecnh b\u00e0y nguy\u00ean t\u1eed kh\u1ed1i l\u00e0 g\u00ec, b\u1ea3ng nguy\u00ean t\u1eed kh\u1ed1i, c\u00e1ch d\u00f9ng \u0111\u1ec3 t\u00ednh ph\u00e2n t\u1eed kh\u1ed1i, ph\u00e2n bi\u1ec7t v\u1edbi s\u1ed1 kh\u1ed1i v\u00e0 b\u00e0i t\u1eadp c\u00f3 l\u1eddi gi\u1ea3i.<\/p>\n

Nguy\u00ean t\u1eed kh\u1ed1i l\u00e0 g\u00ec?<\/h2>\n

Nguy\u00ean t\u1eed kh\u1ed1i l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed9t nguy\u00ean t\u1eed t\u00ednh b\u1eb1ng \u0111\u01a1n v\u1ecb cacbon (vi\u1ebft t\u1eaft l\u00e0 \u0111vC). Ng\u01b0\u1eddi ta quy \u01b0\u1edbc l\u1ea5y 1\/12 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a nguy\u00ean t\u1eed cacbon l\u00e0m \u0111\u01a1n v\u1ecb, g\u1ecdi l\u00e0 m\u1ed9t \u0111\u01a1n v\u1ecb cacbon. Nh\u01b0 v\u1eady, nguy\u00ean t\u1eed kh\u1ed1i cho bi\u1ebft m\u1ed9t nguy\u00ean t\u1eed c\u1ee7a nguy\u00ean t\u1ed1 n\u1eb7ng g\u1ea5p bao nhi\u00eau l\u1ea7n \u0111\u01a1n v\u1ecb cacbon. V\u00ed d\u1ee5, nguy\u00ean t\u1eed oxi c\u00f3 nguy\u00ean t\u1eed kh\u1ed1i l\u00e0 16, ngh\u0129a l\u00e0 n\u1eb7ng g\u1ea5p 16 l\u1ea7n \u0111\u01a1n v\u1ecb cacbon.<\/p>\n

\"nguyen-tu-khoi-1\"<\/p>\n

B\u1ea3ng nguy\u00ean t\u1eed kh\u1ed1i m\u1ed9t s\u1ed1 nguy\u00ean t\u1ed1 th\u01b0\u1eddng g\u1eb7p<\/h2>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Nguy\u00ean t\u1ed1<\/th>\nK\u00fd hi\u1ec7u<\/th>\nNguy\u00ean t\u1eed kh\u1ed1i (\u0111vC)<\/th>\n<\/tr>\n<\/thead>\n
Hi\u0111ro<\/td>\nH<\/td>\n1<\/td>\n<\/tr>\n
Cacbon<\/td>\nC<\/td>\n12<\/td>\n<\/tr>\n
Nit\u01a1<\/td>\nN<\/td>\n14<\/td>\n<\/tr>\n
Oxi<\/td>\nO<\/td>\n16<\/td>\n<\/tr>\n
Natri<\/td>\nNa<\/td>\n23<\/td>\n<\/tr>\n
Magie<\/td>\nMg<\/td>\n24<\/td>\n<\/tr>\n
Nh\u00f4m<\/td>\nAl<\/td>\n27<\/td>\n<\/tr>\n
L\u01b0u hu\u1ef3nh<\/td>\nS<\/td>\n32<\/td>\n<\/tr>\n
Clo<\/td>\nCl<\/td>\n35,5<\/td>\n<\/tr>\n
S\u1eaft<\/td>\nFe<\/td>\n56<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

D\u00f9ng nguy\u00ean t\u1eed kh\u1ed1i \u0111\u1ec3 t\u00ednh ph\u00e2n t\u1eed kh\u1ed1i<\/h2>\n

Ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a m\u1ed9t ch\u1ea5t b\u1eb1ng t\u1ed5ng nguy\u00ean t\u1eed kh\u1ed1i c\u1ee7a t\u1ea5t c\u1ea3 c\u00e1c nguy\u00ean t\u1eed trong ph\u00e2n t\u1eed. V\u00ed d\u1ee5:<\/p>\n

    \n
  • Ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a n\u01b0\u1edbc H\u2082O = 2 \u00d7 1 + 16 = 18 \u0111vC.<\/li>\n
  • Ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a kh\u00ed cacbonic CO\u2082 = 12 + 2 \u00d7 16 = 44 \u0111vC.<\/li>\n
  • Ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a axit sunfuric H\u2082SO\u2084 = 2 \u00d7 1 + 32 + 4 \u00d7 16 = 98 \u0111vC.<\/li>\n<\/ul>\n

    Ph\u00e2n bi\u1ec7t nguy\u00ean t\u1eed kh\u1ed1i v\u00e0 s\u1ed1 kh\u1ed1i<\/h2>\n

    Nhi\u1ec1u h\u1ecdc sinh d\u1ec5 nh\u1ea7m nguy\u00ean t\u1eed kh\u1ed1i v\u1edbi s\u1ed1 kh\u1ed1i. S\u1ed1 kh\u1ed1i (k\u00fd hi\u1ec7u A) l\u00e0 t\u1ed5ng s\u1ed1 proton v\u00e0 s\u1ed1 neutron trong h\u1ea1t nh\u00e2n, lu\u00f4n l\u00e0 s\u1ed1 nguy\u00ean. Trong khi \u0111\u00f3, nguy\u00ean t\u1eed kh\u1ed1i l\u00e0 kh\u1ed1i l\u01b0\u1ee3ng t\u01b0\u01a1ng \u0111\u1ed1i c\u1ee7a nguy\u00ean t\u1eed t\u00ednh theo \u0111\u01a1n v\u1ecb cacbon v\u00e0 c\u00f3 th\u1ec3 l\u00e0 s\u1ed1 th\u1eadp ph\u00e2n (v\u00ed d\u1ee5 clo l\u00e0 35,5 do l\u00e0 gi\u00e1 tr\u1ecb trung b\u00ecnh c\u1ee7a c\u00e1c \u0111\u1ed3ng v\u1ecb). V\u1edbi nhi\u1ec1u nguy\u00ean t\u1ed1, nguy\u00ean t\u1eed kh\u1ed1i x\u1ea5p x\u1ec9 b\u1eb1ng s\u1ed1 kh\u1ed1i, nh\u01b0ng b\u1ea3n ch\u1ea5t hai \u0111\u1ea1i l\u01b0\u1ee3ng n\u00e0y kh\u00e1c nhau.<\/p>\n

    \"nguyen-tu-khoi-2\"<\/p>\n

    M\u1eb9o ghi nh\u1edb nguy\u00ean t\u1eed kh\u1ed1i<\/h2>\n

    \u0110\u1ec3 nh\u1edb nhanh nguy\u00ean t\u1eed kh\u1ed1i, h\u1ecdc sinh th\u01b0\u1eddng d\u00f9ng “b\u00e0i ca nguy\u00ean t\u1eed kh\u1ed1i” \u2013 nh\u1eefng c\u00e2u v\u0103n v\u1ea7n gh\u00e9p t\u00ean nguy\u00ean t\u1ed1 v\u1edbi con s\u1ed1 nguy\u00ean t\u1eed kh\u1ed1i, v\u00ed d\u1ee5 “Hidro l\u00e0 1, m\u01b0\u1eddi hai c\u1ed9t Cacbon, Nit\u01a1 m\u01b0\u1eddi b\u1ed1n tr\u00f2n, Oxi tr\u0103ng m\u01b0\u1eddi s\u00e1u\u2026”. C\u00e1ch h\u1ecdc n\u00e0y gi\u00fap ghi nh\u1edb d\u1ec5 d\u00e0ng v\u00e0 l\u00e2u h\u01a1n.<\/p>\n

    \"nguyen-tu-khoi-3\"<\/p>\n

    B\u00e0i t\u1eadp v\u1eadn d\u1ee5ng c\u00f3 l\u1eddi gi\u1ea3i<\/h2>\n
      \n
    • B\u00e0i 1:<\/strong> T\u00ednh ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a kh\u00ed oxi O\u2082. L\u1eddi gi\u1ea3i:<\/em> = 2 \u00d7 16 = 32 \u0111vC.<\/li>\n
    • B\u00e0i 2:<\/strong> Nguy\u00ean t\u1eed n\u00e0o n\u1eb7ng h\u01a1n, natri (Na) hay oxi (O)? L\u1eddi gi\u1ea3i:<\/em> Na c\u00f3 nguy\u00ean t\u1eed kh\u1ed1i 23, O c\u00f3 16, n\u00ean natri n\u1eb7ng h\u01a1n.<\/li>\n
    • B\u00e0i 3:<\/strong> T\u00ednh ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a canxi cacbonat CaCO\u2083 (Ca = 40). L\u1eddi gi\u1ea3i:<\/em> = 40 + 12 + 3 \u00d7 16 = 100 \u0111vC.<\/li>\n
    • B\u00e0i 4 (tr\u1eafc nghi\u1ec7m):<\/strong> Nguy\u00ean t\u1eed kh\u1ed1i c\u1ee7a cacbon l\u00e0 bao nhi\u00eau? A. 1 B. 12 C. 16 D. 14. \u0110\u00e1p \u00e1n:<\/em> B.<\/li>\n<\/ul>\n

      V\u00ec sao d\u00f9ng \u0111\u01a1n v\u1ecb cacbon?<\/h2>\n

      V\u00ec kh\u1ed1i l\u01b0\u1ee3ng th\u1ef1c c\u1ee7a m\u1ed9t nguy\u00ean t\u1eed v\u00f4 c\u00f9ng nh\u1ecf (c\u1ee1 10\u207b\u00b2\u2074 \u0111\u1ebfn 10\u207b\u00b2\u00b3 gam), r\u1ea5t kh\u00f3 d\u00f9ng trong t\u00ednh to\u00e1n h\u1eb1ng ng\u00e0y, n\u00ean ng\u01b0\u1eddi ta ch\u1ecdn m\u1ed9t \u0111\u01a1n v\u1ecb kh\u1ed1i l\u01b0\u1ee3ng quy \u01b0\u1edbc g\u1ecdi l\u00e0 \u0111\u01a1n v\u1ecb cacbon (\u0111vC). \u0110\u01a1n v\u1ecb n\u00e0y b\u1eb1ng 1\/12 kh\u1ed1i l\u01b0\u1ee3ng nguy\u00ean t\u1eed cacbon, cho ph\u00e9p bi\u1ec3u di\u1ec5n kh\u1ed1i l\u01b0\u1ee3ng c\u00e1c nguy\u00ean t\u1eed b\u1eb1ng nh\u1eefng con s\u1ed1 \u0111\u01a1n gi\u1ea3n, d\u1ec5 so s\u00e1nh v\u00e0 t\u00ednh to\u00e1n (nh\u01b0 H = 1, O = 16). \u0110\u00e2y l\u00e0 l\u00fd do nguy\u00ean t\u1eed kh\u1ed1i \u0111\u01b0\u1ee3c t\u00ednh theo \u0111\u01a1n v\u1ecb cacbon thay v\u00ec gam.<\/p>\n

      \"nguyen-tu-khoi-4\"<\/p>\n

      C\u00e2u h\u1ecfi th\u01b0\u1eddng g\u1eb7p v\u1ec1 nguy\u00ean t\u1eed kh\u1ed1i<\/h2>\n

      Nguy\u00ean t\u1eed kh\u1ed1i l\u00e0 g\u00ec?<\/h3>\n

      L\u00e0 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a m\u1ed9t nguy\u00ean t\u1eed t\u00ednh b\u1eb1ng \u0111\u01a1n v\u1ecb cacbon (\u0111vC).<\/p>\n

      M\u1ed9t \u0111\u01a1n v\u1ecb cacbon b\u1eb1ng bao nhi\u00eau?<\/h3>\n

      B\u1eb1ng 1\/12 kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a nguy\u00ean t\u1eed cacbon.<\/p>\n

      Nguy\u00ean t\u1eed kh\u1ed1i c\u1ee7a oxi l\u00e0 bao nhi\u00eau?<\/h3>\n

      Nguy\u00ean t\u1eed kh\u1ed1i c\u1ee7a oxi (O) l\u00e0 16 \u0111vC.<\/p>\n

      Nguy\u00ean t\u1eed kh\u1ed1i d\u00f9ng \u0111\u1ec3 l\u00e0m g\u00ec?<\/h3>\n

      D\u00f9ng \u0111\u1ec3 so s\u00e1nh \u0111\u1ed9 n\u1eb7ng nh\u1eb9 gi\u1eefa c\u00e1c nguy\u00ean t\u1eed v\u00e0 t\u00ednh ph\u00e2n t\u1eed kh\u1ed1i c\u1ee7a c\u00e1c ch\u1ea5t.<\/p>\n

      Nguy\u00ean t\u1eed kh\u1ed1i v\u00e0 kh\u1ed1i l\u01b0\u1ee3ng mol kh\u00e1c nhau th\u1ebf n\u00e0o?<\/h3>\n

      Ch\u00fang c\u00f3 c\u00f9ng tr\u1ecb s\u1ed1 nh\u01b0ng kh\u00e1c \u0111\u01a1n v\u1ecb: nguy\u00ean t\u1eed kh\u1ed1i t\u00ednh b\u1eb1ng \u0111vC, kh\u1ed1i l\u01b0\u1ee3ng mol t\u00ednh b\u1eb1ng g\/mol.<\/p>\n