K
kobato_2509
$K_2SO_4+BaCl_2\rightarrow 2KCl+BaSO_4$
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$2KCl+H_2SO_4 --->500^0C---2HCl+K_2SO_4$
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$K_2SO_4+BaCl_2\rightarrow 2KCl+BaSO_4$
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$2KCl+H_2SO_4 --->500^0C---2HCl+K_2SO_4$
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$2HCl+Na_2CO_3\rightarrow 2NaCl+H_2O+CO_2$
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$Na_2CO_3 + H_2O +CO_2 \rightarrow 2NaHCO_3$
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$H_2O + CrO_3 \rightarrow H_2CrO_4$$NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O$
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$H_2O + CrO_3 \rightarrow H_2CrO_4$
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$2Mg + CO_2 \to 2MgO + C$ ($Mg$ cháy được trong $CO_2$)$H_2CrO_4+Na_2CO_3\rightarrow Na_2CrO_4+H_2O+CO_2$
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$MgO+H_2O---->Mg(OH)_2$
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$KOH + HCl \rightarrow KCl + H_2O$$h_2o+k----------->koh+h_2$
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$KOH + HCl \rightarrow KCl + H_2O$
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$KCl+AgNO_3\rightarrow AgCl+KNO_3$
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$o_2+h_2--------->h_2o$
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$2H_2O + SO_2 + Br_2 \rightarrow H_2SO_4 + 2HBr$
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$2HBr--t^o--> H_2+Br_2$
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$ H_2O + CO_2 \rightarrow (-C_6H_{10}O_5-)_n + O_2 $
đ/k: Ánh sáng; clorophin ..........