ГДЗ-Химия-9-класс-Габриелян-2002 (9 класс - Габриелян), страница 5

m(H2O) = V(H2O)⋅ρ(H22 fe⋅]fhev ]GZc^_ffZkkmjZkl\hjZkhebPh[s P+2O) + m(CuSO4⋅5H2O) = 5000 ]GZc^_ffZkkmkmevnZlZf_^b[_a\h^gh]hm(CuSO4) = m(CuSO4⋅5H2O)⋅ω \ &X62⋅GH(CuSO4) == m(CuSO4)⋅0U&X62 0U&X62 + 0U+ 2]⋅ + Lh]^ZfZkkh\Zy^hey&X624 \ jZkl\hj_m(CuSO 4 )320ω(CuSO4) =≈0,058 = 5,8 %.=m(h[s )5500]Hl\_l: ω(CuSO4) = 5,8 %.]⋅ ]42<hijhk<gbfZl_evgh jZkkfhljbl_ ko_fm ijbf_g_gby k_jghc dbkehlujbk gZkljb ihiulZcl_kv kZfbhl\_lblvgZwlhl\hijhk§ 23:ahl<hijhkZ0J + N2 = Mg3N2 ²gbljb^fZ]gby0J − H = 0J +1 + H = 1−[KZ + N2 = Ca3N2 ²gbljb^dZevpby&D − H = &D +1 + H = 1−\$O + N2 = 2AlN ²gbljb^Zexfbgby$O − H = $O+1 + H = 1−< ijh^mdlZo j_Zdpbc bhgguc lbi obfbq_kdhc k\yab b bhggZydjbklZeebq_kdZy j_r_ldZGbljb^uf_lZeeh\e_]dh]b^jhebamxlkyk \u^_e_gb_fZffbZdZMg3N2 + 6H2O 3Mg(OH)2↓ + 2NH3↑;<hijhk1 → /L1 → 1+ → 12 → 12 → +121.

N2 + 6Li = 2Li3N; 1 + H → 1 −2. Li3N + 3H2O = 3LiOH + NH3↑; Li – 1 H → Li+;3. 4NH3 H2 = 4NO↑ + 6H2O;4. 2NO + O2 = 2NO2;5. 4NO2 + 2H2O + O2 = 4HNO3.cat<hijhkJZkkfhljbfke_^mxsmxj_Zdpbx2NO + O2 ∅ 2NO2 + Q.Ij_`^_\k_]hwlZj_Zdpbyhdbkebl_evgh\hkklZgh\bl_evgZyhdbke_gb_1 + − H → 1+ ;\hkklZgh\bl_ev  _gb_→ 2O 2− .O 02 + 4e \hkklZgh\ehdbkebl_ev43LZd`_j_Zdpbykbgl_aZ122 y\ey_lkyh[jZlghcj_Zdpb_cl_hgZb^_lh^gh\j_f_ggh\ h[_klhjhguLd\ j_ZdpbbmqZkl\mxl]Zaulhfufh`_fijbf_gblvijbgpbiE_RZl_ev_< gZr_cj_Zdpbbdhebq_kl\hfhev]Zah\ke_\Z[hevr_q_fkijZ\ZLh]^Zijbm\_ebq_gbb^Z\e_gbyjZ\gh\_kb_kf_klblky\ijZ\h< ijyfhcj_Zdpbb\u^_ey_lkyl_iehke_^h\Zl_evghijbmf_gvr_gbbl_fi_jZlmjujZ\gh\_kb_kf_klblky\ijZ\h<hijhkZ1+3 + 5O2 = 4NO↑ + 6H2O1− − H = 1 +2+ H = 2−[1+3 + 3O2 ⇒ 2N2↑ + 6H2O1− − H = 12+ H = 2−.<hijhk>Zgh: m(NH4NO2 ]ωijbf_kb %.GZclb: V(N2), ν(H2O)J_r_gb_GZc^_ffZkkmkheb[_aijbf_k_cm(NH4NO2)=m(NH4NO2ij±P1+4NO2)⋅ωij ]±]⋅0,012=19,76]AZibr_fmjZ\g_gb_j_Zdpbb]W1+ 12 fhev0 ]fhevP ]oe1 fhev9 ]fhev9 ]m]+ 2 fhevF ]fhevP ]GZc^_f912b P+2O):] ²e] ²o e⋅o 19,76 22,4 e912 e64] ²]] ²m ]19,76 ⋅ 36y== 11,115 ]P+22 ]64Lh]^Zν(H2O) = P+ 2 = ]fhev0+ 2] fhevHl\_l: V(N2 eν(H22 fhev44§ 24:ffbZd<hijhk:ffbZd e_]dh k`b`Z_lky ijb h[uqghf ^Z\e_gbb b l_fi_jZlmj_ ±33,4°KZ ijbbkiZj_gbb`b^dh]h]ZaZbahdjm`Zxs_ckj_^uih]ehsZ_lkyfgh]hl_ieZihwlhfmhgijbf_gy_lky\ oheh^bevguomklZgh\dZo:ffbZdhq_gvohjhrhjZkl\hjbf\ \h^_ jZkl\hjZffbZdZgZau\Z_lkygZrZlujgufkibjlhfb bkihevam_lky\ f_^bpbg_2<hijhk<h^hjh^gZyk\yav ²wlhobfbq_kdZyk\yavf_`^mZlhfZfb\h^hjh^Zb ZlhfZfbkbevghwe_dljhhljbpZl_evguowe_f_glh\nlhjdbkehjh^Zahl<h^hjh^gZyk\yavh[jZam_lkyh[uqghf_`^m^\mfykhk_^gbfbfhe_dmeZfbGZijbf_jhgZh[jZam_lkyf_`^mfhe_dmeZfb\h^ukibjlh\nlhjh\h^hjh^ZZffbZdZWlhhq_gvkeZ[Zyk\yav ²ijbf_jgh\ jZakeZ[__dh\Ze_glghc;eZ]h^Zjy _c g_dhlhju_ gbadhfhe_dmeyjgu_ \_s_kl\Z h[jZamxlZkkhpbZlu qlh ijb\h^bl d ih\ur_gbx l_fi_jZlmj ieZ\e_gby bdbi_gby\_s_kl\:ghfZevgh \ukhdb_ l_fi_jZlmju ieZ\e_gby b dbi_gby oZjZdl_jgu^ey\h^u_kebjZkkfZljb\Zlv \h^hjh^gu_kh_^bg_gby9, ]jmiiu<k_\h^hjh^gu_kh_^bg_gby9,]jmiiudjhf_\h^uy\eyxlky]ZaZfbHq_gv\Z`gmxjhevb]jZ_l\h^hjh^gZyk\yav\ fhe_dmeZo\Z`g_crbo^ey`b\uokms_kl\kh_^bg_gbc ²[_edh\b gmde_bgh\uodbkehl3<hijhk>hghjghZdp_ilhjgZyk\yav ²wlhhkh[uc\b^obfbq_kdhc k\yabdhlhjZy\hagbdZ_lg_aZkq_lkiZjb\Zgbyg_iZjguowe_dljhgh\Z aZkq_lk\h[h^ghcwe_dljhgghciZjuh^gh]hZlhfZb k\h[h^ghcyq_cdb^jm]h]h4<hijhk−Kl_i_gvhdbke_gbyZahlZ\ oehjb^_Zffhgby1+4Cl: 1+ &O .−+Kl_i_gvhdbke_gbyZahlZ\ gbljZl_Zffhgby1+4NO3: 1+ 12 .<hijhk3CuO + 2NH3 = N2 + 3Cu + 3H2O.&X + + H = &X 1−− H = 1.>eyhl\_lZbkihevah\Zekymq_[gbdklj>eyhl\_lZbkihevah\Zekymq_[gbdklj4>eyhl\_lZbkihevah\Zekymq_[gbdklj2345<hijhk:ffbZd ijhy\ey_l lhevdh \hkklZgh\bl_evgu_ k\hckl\Z ld ZahlgZoh^blky \ kZfhc gbadhc kl_i_gb hdbke_gby b g_ fh`_l [hevr_ijbgbfZlvwe_dljhgu§ 25KhebZffhgby<hijhkZ1+4)2SO4 + BaCl2 = BaSO4↓ + 2NH4Cl; SO42– + Ba2+ = BaSO4↓.[1+4Cl + AgNO3 = AgCl↓ + NH4NO3; Ag+ + Cl– = AgCl↓.<hijhkZ1+4)2CO3+H2SO4=(NH4)2SO4+H2O+CO2 ↑; CO3–2+2H+=H2O+CO2↑.[1+4)2CO3+2NaOH=Na2CO3+2NH3↑+2H2O; NH4++OH–=NH3↑+H2O.\1+4)2CO3 + CaCl2 = CaCO3↓ + 2NH4Cl; CO32– + Ca2 + = CaCO3↓.W]1+4)2CO3 CO2↑ + 2NH3↑ + H2O↑.<hijhk(NH4)3PO4 ²nhknZlZffhgby(NH4)2HPO4 ²]b^jhnhknZlZffhgbyNH4H2PO4 ²^b]b^jhnhknZlZffhgby(NH4)3PO4 ∅ 3NH4+ + PO43–.(NH4)2HPO4 ∅ 2NH4+ + H+ + PO43–(NH4)H2PO4 ∅ NH4+ + 2H+ + PO43–.<hijhk3241N2 NH3 → NH4Cl → (NH4)2HPO4 →→ NH4NO3.WSFDW1.

N2 + 3H2 - 2NH3↑;2. H3PO4 + 2NH3 = (NH4)2HPO4;3. (NH4)2HPO4 G&O 1+4Cl + H3JH4;4. NH4Cl + AgNO3 = AgCl↓ + NH4NO3.<hijhk>ZghP((NH4)2SO4 d]GZclb91+3), m(NH3)J_r_gb_GZc^_ffZkkh\mx^hexZffbZdZ\ kmevnZl_Zffhgby2 ⋅ 17] / fhev01+ ω1+ 62 (NH3)===0,2576=25,76 %.01+ 62 132] / fhevGZc^_ffZkkmZffbZdZg_h[oh^bfh]h^eyihemq_gbyd](NH4)2SO4:m(NH3) = m((NH4)2SO4)⋅ω(NH3 d]⋅ d]GZc^_fh[t_fZffbZdZ46 ⋅ ]⋅efhev e V(NH3)=ν(NH3)⋅Vν= P1+ ⋅Vν=01+ ]fhevf3.Hl\_l: m(NH3 d]91+3 f3.§ 26Dbkehjh^gu_kh_^bg_gbyZahlZ<hijhk:ahlgZy dbkehlZ g_ h[jZam_l dbkeuo khe_c ld hgZ h^ghhkgh\gZy bbf__l\k_]hh^bgZlhf\h^hjh^Zdhlhjucb hl^Z_lijbh[jZah\Zgbbkheb<hijhkZ&X2+2 + 2HNO3=2H2O + Cu(NO3)2; Cu(OH)2 + 2H+=Cu2+ + 2H2O;[)H2O3 + 6HNO3 = 2Fe(NO3)3 + 3H2O; Fe2O3 + 6H+ = 2Fe3+ + 3H2O.\1D2CO3+2HNO3=2NaNO3 + H2O + CO2↑; CO32– + 2H+=H2O +CO2↑.<hijhkNa2SiO3 + 2HNO3 = 2NaNO3 + H2SiO3↓;SiO32– + 2H+ = H2SiO3↓.<hijhk1) 3Cu + 8HNOjZa[ = 3Cu(NO3)2 + 2NO↑ + 2H2O,&X − H &X 1 H 1.2) Cu + 4HNO3(dhgp) = Cu(NO3)2 + 2NO2↑ + 2H2O&X − H &X 1 H 1<hijhk213→I.

N2 NO2 → NO → HNO3.1. N2 + O2 = 2NO,1 − H = 1 +2 + H = 2−1 + − H = 1 + ..2. 2NO + O2 = 2NO2,2 + H = 2−1 + − H = 1+.3. 4NO2 + O2 + 2H2O = 4HNO3,2 + H = 2−213→ NO2 II. NH3 → NO → HNO3.1. 4NH3 + 5O2 = 4NO + 6H2O,471− − H = 1 +2 + H = 2−.MjZ\g_gbyb kfhljb\ur_<hijhk>Zgh: V(NO2 f3; ω(HNO3) = 68 %.GZclbP+123)J_r_gb_AZibr_fmjZ\g_gb_j_Zdpbbihemq_gbyZahlghcdbkehlu⋅ e12 fhev9ν efhev9 e2 +289,6 e — 252134,4⋅103 e ²o ]o ] ⋅ ⋅ eeo]+12 fhev0 ]fhevP ]d]GZc^_ffZkkmjZkl\hjZZahlghcdbkehlux379 d]m(HNO3) === 555,9 d];0,68ω( HNO 3 )Hl\_l: m(HNO3) = 555,9 d]<hijhk>Zgh: m(NaNO3 ]922 eGZclbPijbf_kbJ_r_gb_AZibr_fmjZ\g_gb_j_Zdpbbo]1D12 fhev0 ]fhevP ]W1D12 ] ²eo ] ²eo ⋅ ]e2 ↑ fhev9 efhev9 e mqbklZy (NaNO3 ]GZc^_ffZkkmijbf_kbPij P1D123)–mqbklZy (NaNO3 ]±] ]Pωij = ij = bebPk_e Hl\_l: ωij = 25%.<hijhkKf\hijhk,,48§ 27Nhknhjb _]hkh_^bg_gby<hijhk1. Na3PO4 ²nhknZlgZljby Na3PO4 = 3Na+ + PO43–;2.

Na2HPO4 ²]b^jhnhknZlgZljby Na2G324 = 2Na+ G+ + PO43–;3. NaH2PO4 ²^b]b^jhnhknZlgZljby 1DG2PO4 = Na+ G+ + PO43–.<hijhk23J→ Mg3P2 →→→ H3PO4 → Ca3(PO4)2.PH3 P2O5 1451. J + 3Mg = Mg3P2,t3 H 30J − H 0J,2. Mg3P2 + 6H2O = 3Mg(OH)2↓ + 2PH3↑.3. 2PH3 + 4O2 = P2O5 + 3H2O,3− − H = 3+2 + H = 2−,4. P2O5 + 3H2O = 2H3PO4;5. 2H3PO4 &DHG2 = Ca3(PO4)2↓ + 6H2H<hijhk3P + 5HNO3 + 2H2O = 3H3PO4 + 5NO↑,3 − H = 3 +.1 + + H = 1 +<hijhk>Zgh: m(Ca3(PO)4 ij ]ωij %, W = 80 %.GZclb: m(P)J_r_gb_GZc^_ffZkkmkheb[_aijbf_kbm(Ca3(PO4)2) = m(Ca3(PO4)2 ij±P&D3(PO4)2 ij⋅ωij]± ] ⋅ ]AZibr_fmjZ\g_gb_j_Zdpbb]&D 32 6L2 & &D6L2 fhev0 ]fhevP ]] ²]] ²o ]o ⋅ ]o]3↑ fhev0 ]fhevP ]m(P)l_hj ]GZc^_ffZkkmnhknhjZdhlhjZyihemqbeZkv\ oh^_j_Zdpbbm(P)ijZdl = m(P)l_hj⋅: ]⋅ ]49Hl\_l: m(P)ijZdl ]<hijhk>ZghP3 ]ωij %,ω(H3PO4) = 80 %.GZclb: mjjZ(H3PO4)J_r_gb_GZc^_ffZkkmnhknhjZ[_aijbf_kbm(P) = m(P ij – m(P ij⋅ωij ] – 31⋅ ]Bah^gh]hfhevnhknhjZfh`ghihemqblvjh\ghh^bgfhevnhknhjghcdbkehluP3P3 29,45]ν(H3PO4)=ν(P)=⇒ν(H3PO4)===0,95 fhev030331]Lh]^ZP+3PO4)=M(H3PO4)⋅ν(H3PO4 ]fhev⋅fhev ]FZkkZjZkl\hjZnhknhjghcdbkehlujZ\gZm(H 3 PO 4 )PjjZ+3PO4) == ] ]ω(H 3 PO 4 )Hl\_lPjjZ+3PO4 ]<hijhk>ZghPjjZ+3PO4 ]ω1(H3PO4) = 5 PJ2H5 ]GZclb: ω2(H3PO4)J_r_gb_AZibr_fmjZ\g_gb_j_Zdpbbihemq_gbynhknhjghcdbkehluG2H J2H5 G3JH4.BamjZ\g_gby\b^ghqlhbafhevJ2H5 ihemqZ_lkyfhevG3JH4.GZc^_ffZkkmnhknhjghcdbkehluihemqb\r_cky\ oh^_j_Zdpbbν(H3PO4) = 2ν(P2O5) = 2⋅P32 ]= 2⋅032 ] fhevfhevm(H3PO4) = ν(H3PO4)⋅M(H3PO4 fhev⋅]fhev ]GZc^_fkdhevdh[uehnhknhjghcdbkehlu\ jZkl\hj_m(H3PO4 PjjZ⋅ω1(H3PO4) = 980⋅ ]GZc^_fgh\mxfZkkh\mx^hexdbkehlu\ jZkl\hj_209,72 + 49ω2(H3PO4) = P+32 h[s == 0,2286.Pj − jZ + P3 2 980 + 152Hl\_l: ω2(H3PO4) = 22,86 %.<hijhk:DhgZg>hce \ k\h_f ijhba\_^_gbb g_ mq_e obfbq_kdbo k\hckl\nhknhjZ >Z\Zcl_ ih^mfZ_f \f_kl_ Nhknhj k\_lblky \ l_fghl_ke_^h\Zl_evgh wlh [_euc nhknhj ld djZkguc nhknhj g_ k\_lblky\hh[s_ >Ze__ [_euc nhknhj hq_gv y^h\bl b bkihevah\Zlv _]h klhev50g_h[uqgufkihkh[hfg_kdhevdhjZa\_jhylgh[ueh[ukeh`ghDjhf_lh]h [_euc nhknhj gZ \ha^mo_ e_]dh hdbkey_lky i_j_oh^y ijb wlhf \kl_i_gvhdbke_gby4P + 5O2 = 2P2O5.DZdlhevdhijhp_kkhdbke_gbyaZdZgqb\Z_lkykh[ZdZi_j_klZ_lk\_lblvkyHdkb^nhknhjZ9hq_gv]b]jhkdhibqgh_\_s_kl\hl_hghijbly]b\Z_ld k_[_\h^m<kihfgbfqlh\ hlju\d_]h\hjbehkvhlmfZg_badhlhjh]h\ukdhqbeh©qm^h\bs_ªLmfZg ²wlhkdhg^_gkbjh\Zggu_iZju\h^uLh]^Zkdhj__\k_]hh[jZah\Z\rbckyP2O5 ih^^_ckl\b_f\eZ]bij_\jZlbeky\ nhknhjgmxdbkehlmJ2H5 G2H⇒G3JH4.Ljm^gh ij_^klZ\blv k_[_ qlh kh[ZdZ h[fZaZggZy y^h\bluf\_s_kl\hfb dbkehlhcfh`_l_s_qlheb[hk^_eZlv§ 28M]e_jh^<hijhk:efZa ²ijhajZqgh_djbklZeebq_kdh_ \_s_kl\hkZfh_l\_j^h_ba\k_o ijbjh^guo \_s_kl\ LZdZy l\_j^hklv h[mkeh\e_gZ hkh[hckljmdlmjhc_]hZlhfghcdjbklZeebq_kdhc j_r_ldb< g_cdZ`^ucZlhfm]e_jh^Z hdjm`_g lZdbfb `_ ZlhfZfb jZkiheh`_ggufb \ \_jrbgZoijZ\bevgh]hl_ljZw^jZDjbklZeeuZefZaZh[uqgh[_kp\_lgu_gh[u\Zxlkbg_]h]hem[h]hdjZkgh]hb q_jgh]hp\_lh\Hgbbf_xlhq_gvkbevguc[e_kd[eZ]h^Zjy\ukhdhck\_lhij_ehfeyxs_cb k\_lhhljZ`Zxs_ckihkh[ghklb:efZaohjhrhijh\h^bll_iehh[eZ^Z_lwe_dljhbaheypbhggufbk\hckl\Zfb5=jZnbl ²l_fghk_jh_`bjgh_gZhsmivdjbklZeebq_kdh_\_s_kl\hk f_lZeebq_kdbf[e_kdhf< hlebqb_hlZefZaZ]jZnblfy]dbcb g_ijhajZqgucohjhrhijh\h^bll_iehb we_dljbq_kdbclhdFy]dhklv]jZnblZh[mkeh\e_gZkehbklhckljmdlmjhcjbk\mq_[gbd_< djbklZeebq_kdhcj_r_ld_]jZnblZZlhfum]e_jh^Ze_`Zsb_\ h^ghciehkdhklbijhqghk\yaZgu\ ijZ\bevgu_r_klbm]hevgbdb656Hl\_l\aylbamq_[gbdZkljHl\_l\aylbamq_[gbdZklj51<hijhkK H2 KH2& − &H − \hkklZgh\bl_ev.2 H 2− − hdbkebl_evWK G2 KG4& + H = &− − hdbkebl_ev.+ − H = ++ − \hkklZgh\bl_ev<hijhkDZj[hg ² wlh h^bg ba i_jbh^h\ iZe_hahckdhc wju < dZj[hg_ijh^he`Zxl jZkijhkljZgylvky kihjh\u_ jZkl_gby l_ iZihjhlgbdbHklZldbih]b[rboiZihjhlgbdh\ f_^e_gghhdbkeyebkv ih^keh_f a_feb\ l_q_gb_ fgh]bo fbeebhgh\ e_l < j_amevlZl_ h[jZah\Zebkv h]jhfgu_aZe_`b dZf_ggh]h m]ey dhlhjuc g_ qlh bgh_ dZd hklZldb ^j_\g_crbojZkl_gbc<hijhkDZj[he_g ²wlhjZagh\b^ghklvZdlb\bjh\Zggh]hm]eyKe_^h\Zl_evghhgh[eZ^Z_lohjhrbfbZ^khj[bjmxsbfbk\hckl\Zfbb bkihevam_lky^eyih]ehs_gbyjZaebqguoaZiZoh\\oheh^bevgbdZo<hijhkK + Fe2O3 = 2Fe KH↑;t& − H = & +)H++ H = )Ht2) PbO2 + 2C = Pb + 2CO↑;3E + + H = 3E & − H = &+.<hijhk>ZghPK ij ]9&22 e: %.GZclb: ωijJ_r_gb_GZc^_fh[t_fm]e_dbkeh]h]ZaZdhlhjuc^he`_g[ueihemqblvkyV(CO2)l_hj =9&2 = e:AZibr_fmjZ\g_gb_j_ZdpbbeWK H2 KH2.Ba mjZ\g_gby j_Zdpbb \b^gh qlh ba h^gh]h fhev m]ey ihemqZ_lkyjh\ghh^bgfhevm]e_dbkeh]h ]ZaZAgZqbl52νK νKH2) = ee fhevfhevVνLh]^ZP& νK⋅0K fhev⋅]fhev ]Pij P& ij±PK ]±] ]m(ij)2]Lh]^Zωij == 0,25 = 25 %.m(C + ij)8]Hl\_l: ωij %.<hijhkK KH2 KHV(CO 2 ) l_hj=& − H = &+ − \hkklZgh\bl_ev .K + + H = K + − hdbkebl_ev<hijhkK→ CaC2 → C2H2 → CO2 → CO1234t1.