[問題] 普化的小小題目

[finalpoet (finalpoet)]

1) Two solutions, initially at 24.60°C, are mixed in a coffee cup calorimeter
(Ccal = 15.5 J/°C).  When a 100.0 mL volume of 0.100 M AgNO3 solution is
mixed with a 100.0 mL sample of 0.200 M NaCl solution, the temperature in the
calorimeter rises to 25.30°C.  Determine the ΔH。rxn  for the reaction as written
below.  Assume that the density and heat capacity of the solutions is the
same as that of water.

ΔH。rxn =?

A) -35 kJ
B) -69 kJ
C) -250 kJ
D) -16 kJ
E) -140 kJ


2)What pressure would a gas mixture in a 10.0 L tank exert if it were
composed of 48.5 g He and 94.6 g CO2 at 398 K?

A) 39.6 atm
B) 7.02 atm
C) 32.6 atm
D) 46.6 atm
E) 58.7 atm

3) According to the following reaction, what amount of Al2O3 remains when
20.00 g of Al2O3 and 2.00 g of H2O are reacted?  A few of the molar masses
are as follows:  Al2O3 = 101.96 g/mol, H2O = 18.02 g/mol.

A) 28.33 g
B) 14.00 g
C) 8.33 g
D) 19.78 g
E) 11.67 g


Thanks!!!


--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 203.71.94.31
推 ellenroy:第三題題目沒打錯嗎?哪裡需要Al2S3分子量?
※ 編輯: finalpoet       來自: 203.71.94.31         (05/19 23:18)
→ finalpoet:已修正了，謝謝
推 cen71:第一題算不出來耶~我怎麼算都是59.63KJ
→ cen71:第二題是46.6 用PV=nRT算出兩個P 在相加得解
→ cen71:第三題氧化鋁應該是不溶於水~我可能會選D吧 用猜的><
→ finalpoet:59.63KJ怎麼算出來的阿?第二題感謝囉
→ finalpoet:第三題我覺得是：Al2O3+6H2O→2Al(OH)3+3H2O
推 cen71:如果第三題反應式如上，那剩餘應該是18.1g的氧化鋁
→ cen71:第一題的話..反應係數比為1:1 然後從給的體積跟濃度可求mol
→ cen71:兩個mol數分別為 0.01跟0.02 所以AgNO3是限量
→ cen71:然後用E=(Cc+ms)*ΔT  ms為溶液重跟比熱 題目說跟水一樣
→ cen71:所以ms代 m = 200g (200ml的溶液密度為1) s = 4.18 J/gC
→ cen71:可以算出E= (15.5 + (200*4.18))*0.07 =59.605 為0.01mol反
→ cen71:應熱 所以標準反應焓應為 59.605 / 0.01 =5960.5J =59.6KJ
→ cen71:我不知道我的想法對不對..不過跟答案不一樣是真的 = =
推 cen71:第三題的話就算出兩個mol數 分別為0.196 跟 0.111
→ cen71:而6mol的水跟1mol的氧化鋁作用 所以0.111/6 = 反應的氧化鋁
→ cen71:的mol數 用0.196 - (0.111/6) 得剩餘的氧化鋁mol數
→ cen71:再反算回去就可以求得剩餘氧化鋁的克數
→ a12henry:反應應該不是長那樣 要放鹼在裡面才會變成Al(OH)3
→ finalpoet:感謝C大的解題，所以第三題的條件下是無法反應的囉
推 cen71:依溶解度法則..Al2O3是不溶 應該把他當難溶 D的機會比較大