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    Ce vrei sa afli? Scrie INTREBAREA

    1. Asked: ianuarie 17, 2021In: Chimie

      reactii de aditie 1-butena si 2-butena

      papipapicuisine

      papipapicuisine

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      papipapicuisine
      Added an answer on ianuarie 17, 2021 at 3:38 pm

      CH3 –CH2-HC = CH2 + H2 → CH3–CH2-CH2-CH3 1-butenă hidrogen n-butan CH3 –HC = CH -CH3 + H2 → CH3–CH2-CH2-CH3 2-butenă hidrogen n-butan CH3-CH-HC = CH2 + Cl2 → CH3-CH2-CHCl –CH2Cl 1-butenă clor 1,2 diclorbutan CH3-HC = CH – CH3 + Cl2 → CH3-CHCl - CHCl –CH3 2-butenă clor 2,3 diclorbutan CH3-CH2-HC = CHRead more

      CH3 –CH2-HC = CH2 + H2 → CH3–CH2-CH2-CH3
      1-butenă hidrogen n-butan
      CH3 –HC = CH -CH3 + H2 → CH3–CH2-CH2-CH3
      2-butenă hidrogen n-butan
      CH3-CH-HC = CH2 + Cl2 → CH3-CH2-CHCl –CH2Cl
      1-butenă clor 1,2 diclorbutan
      CH3-HC = CH – CH3 + Cl2 → CH3-CHCl – CHCl –CH3
      2-butenă clor 2,3 diclorbutan
      CH3-CH2-HC = CH2 + H-Cl → H3C –CH2- CHCl –CH3
      1 -butenă acid clorhidric clorură de secbutil – regula lui Markovnicov
      H3C-HC = CH-CH3 + H-Cl → H3C –CH2- CHCl –CH3
      2 -butenă acid clorhidric clorură de secbutil – regula lui Markovnicov – nu e necesară
      CH3-CH2-HC = CH2 + H-OH → H3C –CH2- CH(OH) –CH3
      1 -butenă apă alcool secbutilic – regula lui Markovnicov
      H3C-HC = CH-CH3 + H-OH → H3C –CH2- CH(OH) –CH3
      2 -butenă apă alcool secbutilic – regula lui Markovnicov-nu e necesară

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    2. Asked: ianuarie 17, 2021In: Chimie

      Afla numarul de moli de atomi de fosfor conținuți în 93g fosfor

      papipapicuisine

      papipapicuisine

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      papipapicuisine
      Added an answer on ianuarie 17, 2021 at 3:33 pm

      Știm că un mol de fosfor are 31g.Atunci 93g o sa aibă 3 moli

      Știm că un mol de fosfor are 31g.Atunci 93g o sa aibă 3 moli

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    3. Asked: ianuarie 17, 2021In: Chimie

      Un element chimic are 11 protoni și 12 neutroni

      papipapicuisine

      papipapicuisine

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      papipapicuisine
      Added an answer on ianuarie 17, 2021 at 3:29 pm

      .

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    4. Asked: ianuarie 3, 2021In: Chimie

      Să se determine natura agresivităţii şi gradul acesteia în cazul unei ape care conține:

      papipapicuisine

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      papipapicuisine
      Added an answer on ianuarie 17, 2021 at 3:15 pm

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    5. Asked: ianuarie 3, 2021In: Chimie

      Ce concentratie a avut soluția de amoniac?

      papipapicuisine

      papipapicuisine

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      papipapicuisine
      Added an answer on ianuarie 17, 2021 at 3:14 pm

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    6. Asked: ianuarie 3, 2021In: Chimie

      Acidul sulfuric de concentrație 96% are densitatea 1.84 g/cm². Să se calculeze concentrația

      papipapicuisine

      papipapicuisine

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      papipapicuisine
      Added an answer on ianuarie 17, 2021 at 3:14 pm

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    7. Asked: ianuarie 17, 2021In: Chimie

      Radicalii divalenti proveniți de la etan și propan sunt?

      papipapicuisine

      papipapicuisine

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      papipapicuisine
      Added an answer on ianuarie 17, 2021 at 3:11 pm

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    8. Asked: ianuarie 17, 2021In: Chimie

      Scrie ecuații chimice ale acidului acetic (acidul etanoic) cu:

      papipapicuisine

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      papipapicuisine
      Added an answer on ianuarie 17, 2021 at 2:44 pm

      CH3COOH + H2O => C H 3 C O O - + H 3 O + 2CH3COOH + Zn=> Zn (CH3COO)2 + H2 2CH3COOH+ CaO=> Ca(CH3COO)2+ H2O CH3COOH + NaOH=> CH3COONa + H2O

      CH3COOH + H2O => C H 3 C O O – + H 3 O +

      2CH3COOH + Zn=> Zn (CH3COO)2 + H2

      2CH3COOH+ CaO=> Ca(CH3COO)2+ H2O

      CH3COOH + NaOH=> CH3COONa + H2O

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    9. Asked: ianuarie 17, 2021In: Chimie

      Poate reactiona cu maxim 2 moli de clorura de acetil per mol de substanta:

      papipapicuisine

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      papipapicuisine
      Added an answer on ianuarie 17, 2021 at 2:33 pm

      Raspuns corect: A Grupa OH din serina reactioneaza cu clorura de acetil si se formeaza un ester si grupa NH2 libera din glicina, formand o amina acilata.

      Raspuns corect: A
      Grupa OH din serina reactioneaza cu clorura de acetil si se formeaza un ester si grupa NH2 libera din glicina, formand o amina acilata.

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    10. Asked: ianuarie 3, 2021In: Chimie

      Câţi moli de H2SO4 se găsesc in 20 t soluţie de H2SO4 de concentrație 49% ?​

      papipapicuisine

      papipapicuisine

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      papipapicuisine
      Added an answer on ianuarie 3, 2021 at 6:40 pm

      20 t= 20000 kg ms H2SO4= 20000kg c= 49 % c%=(md/ms) *100 n= m : M moli H2SO4 = md md= (c x ms) : 100 md= (49 x 20000) : 100= 9800 kg H2SO4 M acid sulfuric=2*1+32+4*16=98 n= 9800kg : 98 kg/kmoli= 100 moli

      20 t= 20000 kg

      ms H2SO4= 20000kg

      c= 49 %
      c%=(md/ms) *100
      n= m : M

      moli H2SO4 = md

      md= (c x ms) : 100

      md= (49 x 20000) : 100= 9800 kg H2SO4

      M acid sulfuric=2*1+32+4*16=98

      n= 9800kg : 98 kg/kmoli= 100 moli

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