Skip to content

  • Home
  • Mathematics
  • English
  • History
  • Chemistry
  • Biology
  • Law
  • Medicine
  • Business
  • Toggle search form

Hòa tan hoàn toàn 1,6g CUO trong 100g dung dịch H2SO4 20%.Tính nồng độ phần trăm của các chất có trong dung dịch

Posted on October 23, 2021 By Ioldelk123 6 Comments on Hòa tan hoàn toàn 1,6g CUO trong 100g dung dịch H2SO4 206.Tính nồng độ phần trăm của các chất có trong dung dịch

hòa tan hoàn toàn 1,6g CUO trong 100g dung dịch H2SO4 20%.Tính nồng độ phần trăm của các chất có trong dung dịch thu được

Chemistry

Post navigation

Previous Post: GEOMETRY HELP PLEASE
Next Post: Direct materials costs for an automobile manufacturer would include all of the following:.a. tires b. employees who work on

Comments (6) on “Hòa tan hoàn toàn 1,6g CUO trong 100g dung dịch H2SO4 20%.Tính nồng độ phần trăm của các chất có trong dung dịch”

  1. tatumleigh04 says:
    October 23, 2021 at 6:47 am

    Mynau ne men seni tsunbedim

    Reply
  2. hkcapricorn7705 says:
    October 23, 2021 at 7:03 am

    a) PTHH: CuO + H2SO4 ---> CuSO4 + H2O  

    b)  

    n Cu = 1,6 / 80 = 0,02 mol  

    m H2SO4 = 20 . 100 / 100 = 20 g  

    => n H2SO4 = 20 / 98 = 0,204 mol  

    TPT:  

    1 mol : 1 mol  

    0,02 mol : 0,204 mol  

    => Tỉ lệ: 0,02/1 < 0,204/1  

    => H2SO4 dư, tính toán theo CuO  

    m dd sau p/ư = m dd H2SO4 + m CuO = 100 + 1,6 = 101,6 g  

    TPT: n CuSO4 = n CuO = 0,02 mol  

    => m CuSO4 = 0,02 . 160 = 3,2 g  

    => C% CuSO4 = 3,2 / 101,6 . 100% = 3,15%  

    n H2SO4 dư = 0,204 - 0,02 = 0,182 mol  

    => m H2SO4 dư = 0,182 . 98 =17,836 g  

    => C% H2SO4 = 17,836 / 101,6 . 100% = 17,83%

    Explanation:

    Reply
  3. marcgotbox says:
    October 23, 2021 at 10:17 am

    I need it translated in English!

    Reply
  4. Expert says:
    October 23, 2021 at 10:18 am

    a electronegativity decreases beacuse the size of nucleus is bigger.

    Reply
  5. Expert says:
    October 23, 2021 at 1:08 pm

    8

    explanation:

    each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

    mark brainliest and have a great day!

    Reply
  6. dondre54 says:
    October 24, 2021 at 5:06 am

    Step-by-step explanation:

    Reply

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Categories

  • Advanced Placement (AP)
  • Arts
  • Biology
  • Business
  • Chemistry
  • Computers and Technology
  • Engineering
  • English
  • French
  • Geography
  • German
  • Health
  • History
  • Law
  • Mathematics
  • Medicine
  • Physics
  • SAT
  • Social Studies
  • Spanish
  • World Languages

© 2021 studyqas

Powered by PressBook WordPress theme