Ta có: \(M_{R_2O}=\dfrac{m}{n}=\dfrac{15,59}{0,25}=62,36\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow2R+16=62,36\)
\(\Rightarrow R=23,18\)
\(\Rightarrow R\) là \(Na\) .
CTHH: \(Na_2O\)
Ta có: \(M_{R_2O}=\dfrac{m}{n}=\dfrac{15,59}{0,25}=62,36\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow2R+16=62,36\)
\(\Rightarrow R=23,18\)
\(\Rightarrow R\) là \(Na\) .
CTHH: \(Na_2O\)
lap cong thuc hoa hoc cua chat biet khoi luong cua luu huynh 12,8 khoi luong oxi 14,8
viet cong thuc hoa hoc va tinh phan tu khoi cua cac hop chat sau
1 dong(ll)sunfat,biet phan tu co 1 Cu,1S,4O lien ket voi nhau
2 axit sunfuric,biet phan tu co 2H 1S 4O lien ket voi nhau
3 muoi an,biet phan tu co 1Na,1Cl lien ket voi nhau
4 sat(ll)photphat , biet phan tu gom 3Fe 2 nhom PO4 lien ket voi nhau
giai nhanh giup em
mot hop chat gom 2 nguyen to la X&O, biet X co hoa tri III &X chiem 70% ve KL phan tu cua hop chat . Biet phan tu cua hop chat nang gap 5 lan phan tu khi Oxi . Xac dinh nguyen to hoc hoc X.
BAI NAY CO TRONG SBT HOA NHA CAC BAN ;BAI 9.8 do
hop chat A tao boi 1 nto la nito va oxi . Nguoi ta xac dinh dc rang , trong ti le khoi luong giua 2 nto trong A = Mn/Mo =7/2
Viet CTHH va tinh PTK cua A
xac dinh cong thuc hoa hoc cua nhom oxit , biet ti le khoi cua 2 nguyen to nhom va oxit bang 4,5:4
lap cong thuc hoa hoc cua chat biet khoi luong cua luu huynh 12,8 khoi luong oxi 1,8
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trong hop chat RO,oxi chiem 19,75%ve khoi luong.cong thuc hoa hoc RO la
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phan tu X co phan tu khoi gap 2 lan phan tu khoi cua Oxi ,X tao nen tu 2nguyen to S va O.Xac dinh cong thuc hoa hoc cua X
Moi nguoi giup mik nha
Minh thanks moi nguoi nhieu
Khi dot nong 1 gam magie ket hop voi 2,96 gam clo tao ra hop chat magie clorua. Hay tim CTHH cua hop chat magie clorua, biet phan tu cua hop chat chi co 1 nguyen tu magie.