24 bít

Vì mỗi ký tự cần 8 bít 

_ có 3 ký tự: ( - )  ( 3 )  ( 0 )  =====> 8 x 3 = 24bit

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n_MgCO3 = \(\frac{m}{M}\)= \(\frac{8,4}{84}\)= 0,1 (mol)

Khi cho MgCO₃ vào HCl, ta có PTHH:

a. MgCO₃ + 2HCl \(\rightarrow\)MgCl₂ + CO₂\(\uparrow\)+ H₂O

        0,1 \(\rightarrow\)0,2      :           0,1  :    0,1     : 0,1            (mol)

b. C%_HCl = \(\frac{mt}{md}\). 100% = \(\frac{36,5.0,2}{146}\).100% = 5 %

c. m_ddsau = m_MgCO3 + m_HCl - m_CO2 = 8,4 + 146 - 44.0,1 = 150 (g)

C%_MgCl2 = \(\frac{mt}{md}\).100% = \(\frac{0,1.95}{150}\).100% \(\approx\) 6,33 %

Số học sinh chỉ giỏi Toán là:

20-10=10(bạn)

Số học sinh chỉ giỏi Lý là:

20-10=10(bạn)

Số học sinh chỉ giỏi Hóa là:

45-10-10=25(bạn)

giúp em với ạ

Ngày đầu tiên sên bò lên được 10m, đêm lại tụt xuống 9 m. Như vậy sau một ngàyđêm, sên bò lên được:   10 - 9 = 1 (m)
    Ngày cuối cùng sên bò lên 10m để được lên tới đỉnh thì sên phải ở mét thứ 10.
    Để sên bò lên và tụt xuống ở đúng mét thứ 10 thì phải mất:     1 x 10 = 10 (ngày đêm)
    Như vậy sên bò lên tới đỉnh phải mất 11  (10+1=11) ngày 10 đêm.

Ngày đầu tiên sên bò lên được 10 m, đêm lại tụt xuống 9 m. Như vậy sau một ngày đêm, sên bò lên được:

10 - 9 = 1 (m)

 Ngày cuối cùng sên bò lên 20m để được lên tới đỉnh thì sên phải ở mét thứ 20

 Để sên bò lên và tụt xuống ở đúng mét thứ 10 thì phải mất:

20 . 1 = 20 (ngày)

Đáp án A

Theo giả thiết đề bài cho, ta có biểu đồ Ven:

Dựa vào biểu đồ Ven ta thấy:

Số học sinh chỉ giỏi Toán và Lý (không giỏi Hóa) là: 6−3=3 (em)

Số học sinh chỉ giỏi Toán và Hóa (không giỏi Lý) là: 4−3=1 (em)

Số học sinh chỉ giỏi Lý và Hóa (không giỏi Toán) là: 5−3=2 (em)

Số học sinh chỉ giỏi một môn Toán là: 10−3−3−1=3 (em)

Số học sinh chỉ giỏi một môn Lý là: 10−3−3−2=2 (em)

Số học sinh chỉ giỏi một môn Hóa là: 11−1−3−2=5 (em)

Số học sinh giỏi ít nhất một trong ba môn là:

3+2+5+1+2+3+3=19 (em)

A