A
分析:A、過氧化鈉由鈉離子與過氧根離子構(gòu)成,根據(jù)n=
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計(jì)算過氧化鈉的物質(zhì)的量,再根據(jù)N=nN
A計(jì)算離子總數(shù).
B、根據(jù)n=
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計(jì)算T
2O的物質(zhì)的量,每個(gè)T
2O分子含有10個(gè)質(zhì)子,再根據(jù)N=nN
A計(jì)算質(zhì)子總數(shù).
C、標(biāo)準(zhǔn)狀況下,三氧化硫是固體.
D、根據(jù)n=
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計(jì)算乙烯的物質(zhì)的量,每個(gè)乙烯分子含有6個(gè)共價(jià)鍵,再根據(jù)N=nN
A計(jì)算共價(jià)鍵總數(shù).
解答:A、7.8gNa
2O
2的物質(zhì)的量為
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=0.1mol,含有的離子總數(shù)為0.1mol×3×N
Amol
-1=0.3N
A,故A正確;
B、2.0gT
2O的物質(zhì)的量為
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=
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mol,每個(gè)T
2O分子含有10個(gè)質(zhì)子,含有質(zhì)子總數(shù)為
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mol×10×N
Amol
-1=
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N
A,故B錯(cuò)誤;
C、0.5N
A個(gè)SO
3分子的物質(zhì)的量為0.5mol,標(biāo)準(zhǔn)狀況下,三氧化硫是固體,不能使用氣體摩爾體積,故C錯(cuò)誤;
D、標(biāo)準(zhǔn)狀況下,22.4L乙烯分子的物質(zhì)的量為1mol,每個(gè)乙烯分子含有6個(gè)共價(jià)鍵,含有共價(jià)鍵總數(shù)為1mol×6×N
Amol
-1=6N
A,故D錯(cuò)誤.
故選:A.
點(diǎn)評:考查常用化學(xué)計(jì)量有關(guān)計(jì)算,難度不大,靈活公式運(yùn)用,注意三氧化硫標(biāo)準(zhǔn)狀況下為固體,通常情況為液體.