Together creative people from all over the world. The prizes are presented Open Eurasia is an international creative competition, which bringsat the Open Eurasian Book Forum & Literature FestivalThis international contest includes a festival and forum. It invites creativeDoes, writers, artists and film directors from the Eurasia region and allover the world. Many writers. poets, translators and artists take part in thecompetition to become a winner.Contest Open Eurasia has been held since 2012. Contest is organizedThe Fesóval is held in different places. For example, in 2018, the festivalwas held in Thailand, and before it took place in Bishkek, London, AlmatyThe winners can publish their books in London. Thanks to the prizes.the authors can show their books in the Open Book Forum Eurasian &Literature Festival and in other places in Europe and Asia!One of the finalists of the Contest in 2019 was a historical novel byan Uzbek author Shahodat Ulug. She has introduced her novel "JayhunEpkinlar (Jayhun Breath" Galina Dolgaya was the first who won the firstonce in the history of the Contest. She lives in Tashkent. She has publisheda number of novels and poems. Her novel "The Gods of the Middle World"won the first prize at the Contest 2012 Literature Festival and it is her first7) Who are the other winners from Uzbekistan?Open Eurasia 2019 Contest Announced!" "99work in English перевод на русский
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2) а
3) NH3 + HCl = NH4Cl
NH3 + H(+) + Cl(-) = NH4(+)+Cl(-) полное ионное
NH3 + H(+) = NH4(+) сокращённое ионное
NH4Cl + AgNO3 = NH4NO3 + AgCl(осадок)
NH4(+) + Cl(-) + Ag(+) + NO3(-) = NH4(+) + NO3(-) + AgCl(осадок) полное ионное
Cl(-) + Ag(+) = AgCl(осадок) сокращённое ионное
NH4NO3 + NaOH = NH3(газ) + NaNO3 + H2O
NH4(+) + NO3(-) + Na(+) + OH(-) = NH3(газ) + Na(+) + NO3(-) + H2O полное ионное
NH4(+) + OH(-) = NH3(газ) + H2O сокращённое ионное
4) S + O2 = SO2
n(O2) = n(S) = 2моль V(O2) = n(O2)*Vмоль = 2моль*22,4л/моль=44,6л
5) 2NH4Cl + Ca(OH)2 = 2NH3(газ) + СaCl2 + 2H2O
n(NH4Cl)=m/M=42/53,5=0,785моль
n(NH3,теор)=n(NH4Cl)=0,785моль V(NH3,теор)=n(NH3,теор)*Vмоль=0,785моль*22,4л/моль=17,585л
w=V(NH3)*100%/V(NH3,теор)=11,6л*100%/17,585л=65,965%
ответ: объёмный выход аммиака равен 65,965%.