найти массовую долю элементов (всех) в соединених?
P2O5
HNo3
AsH3
задачка
Мr(P2O5) = 2Аr(P) + 5Аr(O);Мr(P2O5) = 2 • 31 + 5 • 16 = 142;W(P) = 62/142 = 0, 44 = 44 %
W (O) = 80/142 = 0,56 = 56 %
Мr(HNO3) = Аr(H) + Аr(N) + 3Аr(O) ;Мr(HNO3) = 1 + 14 + 3 • 16 = 63;W(H) = 1/63 = 0, 016 = 1,6 %
W (N) = 14/63 = 0, 22 = 22 %
W (O) = 48/63 = 0, 76 = 76%
Мr(AsH3) = Аr(As) + 3Аr(H) ;Мr(AsH3) = 75 + 3 • 1 = 78;W(As) = 75/78 = 0, 96 = 96 %
W (H) = 3/78 = 0, 038 = 3,8 %
№1P2O5 P=31*2=62, O=16*5=80
80+62=142
142----100%
80------x
x=(80*100)/142=56,3%(O)
100-56,3...=43.66 т.е.44%(P)
ответ:P=43,66%
O=56,3%
№2 HNO3 H=1, N=14,O=16*3=48
1+14+48=63
63----100%
1-------x
x=(100*1)/63=1,58% т.е. 1,6%(H)
63-1=62
100%-1,58...=98,41%
62------98,41%
14-------x
x=(14*98,41)/62=22,22%(N)
98,41-22,22=76,19%(O)
ответ:H=1,6% или 1,58%
N=22,22%
O=76,19%
№3 AsH3 As=75, H=1*3=3
75+3=78
78-----100%
75-----x
x=(75*100)/78=96,15%
100-96,15...=3,84%
ответ:As=96,15%
H=3,84%
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Answers & Comments
Мr(P2O5) = 2Аr(P) + 5Аr(O);
Мr(P2O5) = 2 • 31 + 5 • 16 = 142;
W(P) = 62/142 = 0, 44 = 44 %
W (O) = 80/142 = 0,56 = 56 %
Мr(HNO3) = Аr(H) + Аr(N) + 3Аr(O) ;
Мr(HNO3) = 1 + 14 + 3 • 16 = 63;
W(H) = 1/63 = 0, 016 = 1,6 %
W (N) = 14/63 = 0, 22 = 22 %
W (O) = 48/63 = 0, 76 = 76%
Мr(AsH3) = Аr(As) + 3Аr(H) ;
Мr(AsH3) = 75 + 3 • 1 = 78;
W(As) = 75/78 = 0, 96 = 96 %
W (H) = 3/78 = 0, 038 = 3,8 %
№1P2O5 P=31*2=62, O=16*5=80
80+62=142
142----100%
80------x
x=(80*100)/142=56,3%(O)
100-56,3...=43.66 т.е.44%(P)
ответ:P=43,66%
O=56,3%
№2 HNO3 H=1, N=14,O=16*3=48
1+14+48=63
63----100%
1-------x
x=(100*1)/63=1,58% т.е. 1,6%(H)
63-1=62
100%-1,58...=98,41%
62------98,41%
14-------x
x=(14*98,41)/62=22,22%(N)
98,41-22,22=76,19%(O)
ответ:H=1,6% или 1,58%
N=22,22%
O=76,19%
№3 AsH3 As=75, H=1*3=3
75+3=78
78-----100%
75-----x
x=(75*100)/78=96,15%
100-96,15...=3,84%
ответ:As=96,15%
H=3,84%