SAT2化學(xué)題型之配對(duì)題(Matching questions in Part A)

　　在每一道SAT2化學(xué)配對(duì)題型中，給出五個(gè)選項(xiàng)來(lái)回答所有問(wèn)題。選項(xiàng)可能是說(shuō)明性的文字，圖畫(huà)，圖表，實(shí)驗(yàn)結(jié)果，方程式等。有的問(wèn)題只需要簡(jiǎn)單的回憶一下知識(shí)，有的問(wèn)題需要對(duì)所給信息進(jìn)行定性或定量的分析來(lái)總結(jié)出答案。

　　SAT2化學(xué)配對(duì)題型的說(shuō)明特別強(qiáng)調(diào)每個(gè)選項(xiàng)可以使用一次，多次或者不使用。

　　PART A

　　Directions: Every set of the given choices below refers to the numbered statements or formulas immediately following it. Choose the one lettered choice that best fits each statement or formula and then fill in the corresponding oval on the answer sheet. Each choice may be used once, more than once, or not at all in each set.

　　A部分

　　說(shuō)明：前面的一組選項(xiàng)是緊跟其后的幾個(gè)問(wèn)題的待選答案。選擇出問(wèn)題的最佳選項(xiàng)，然后涂在答題紙相應(yīng)位置上。每個(gè)選項(xiàng)可以使用一次，多次或者不使用。

　　例題

　　Questions 1-3 refer to the following aqueous solutions.

　　(A) 0.1 M HCl

　　(B) 0.1 M NaCl

　　(C) 0.1 M HC2H3O2

　　(D) 0.1 M CH 3 OH

　　(E) 0.1 M KOH

　　1. Is weakly acidic

　　2. Has the highest pH

　　3. Reacts with an equal volume of 0.05 M Ba(OH)2 to form a solution with pH = 7

　　KEY：(C) (E) (A)

　　SAT2化學(xué)題型之判斷題 (True/false and relationship questions in Part B)

　　在實(shí)際的SAT2考試中，這種類型的問(wèn)題要求把答案寫(xiě)在標(biāo)有“化學(xué)”的專用的答題紙上。第二種類型的題目從101道題目開(kāi)始。每道題目的第一欄有一種表述，另一邊的第二欄有另一種表述。你的首要任務(wù)是判斷這些表述是否正確，并在答題紙上相應(yīng)位置選擇T或F;然后運(yùn)用推理能力和對(duì)題目的理解判斷兩種表述是否存在因果關(guān)系。所以說(shuō)小編提醒大家注意在備考中一定要慢慢總結(jié)SAT2化學(xué)原理，重視SAT2考試中對(duì)于詞匯的考察。

　　Part B

　　Directions: Every question below contains two statements, Ⅰin the left-hand column and Ⅱ in the right-hand column. For each question, decide if statementⅠis true or false and if statementⅡis true or false and fill in the corresponding T or F ovals on your answer sheet. Fill in oval CE only if state Ⅱis a correct explanation of statement Ⅰ.

　　說(shuō)明：下面的每個(gè)問(wèn)題包括兩種表述，左欄的Ⅰ和右欄的Ⅱ。對(duì)于每一個(gè)問(wèn)題，判斷Ⅰ和Ⅱ的表述是否正確，并在答題紙上相應(yīng)位置選擇T或F。當(dāng)Ⅱ是Ⅰ的正確解釋時(shí)填涂CE。

　　Sample Answer Grid:

　　CHEMISTRY* Fill in oval CE if only Ⅱis a correct explanation of Ⅰ.

　　EXAMPLE

　　The reaction that takes place is

　�、� 　　 　�、�

　　101.　　When 2 liters of oxygen gas react completely with 2 liters of hydrogen gas, the limiting factor is the volume of the oxygen .　　BECAUSE　　The coefficients in the balanced equation of a gaseous reaction give the volume relationship of the reacting gases.

　�、瘛　� 　�、�

　　101.　　2升氧氣與2升氫氣充分反應(yīng)，限制因素是氧氣的量因?yàn)榕淦降姆磻?yīng)方程式的各個(gè)反應(yīng)物的系數(shù)表明了參加反應(yīng)的氣體之間量的關(guān)系

　　反應(yīng)的化學(xué)方程式為

　　2H2 + O2 → 2H2O

　　The coefficient of this gaseous reaction show that 2L of hydrogen react with 1L of oxygen, leaving 1L of unreacted oxygen. The limiting factor is the quantity of hydrogen.

　　氣體反應(yīng)的系數(shù)表明2L氫氣與1L氧氣反應(yīng)，剩余1L氧氣未參加反應(yīng)。 限制因素應(yīng)是氫氣的量。

　　The ability to solve this quantitative relationship shows that statement I is not true. However, statement II does give a true statement of the relationship of coefficients in a balanced equation of gaseous chemical reaction. Therefore, you should fill F in I and T in II.

　　定量結(jié)果表明I 是不正確的。但是，II 所表述的.配平的反應(yīng)方程式的各個(gè)反應(yīng)物的系數(shù)表明了參加反應(yīng)的氣體之間量的關(guān)系是對(duì)的。因此，在I中選擇F，II中選擇T。

　　SAT2化學(xué)題型之多項(xiàng)選擇題(General multiple-choice question in Part C)

　　這個(gè)SAT2化學(xué)題型通常為問(wèn)題或者不完整表述，帶有五個(gè)備選答案，你必須從中選出最佳答案。在有些題目中，要求選出不恰當(dāng)?shù)拇鸢�。這種問(wèn)題包括大寫(xiě)的單詞,例如NOT, LEAST, EXCEPT。

　　在有的問(wèn)題中，會(huì)要求在圖片，圖表，數(shù)學(xué)表達(dá)式和文字表述之間建立聯(lián)系。解決方法包括對(duì)已知信息的正確解讀來(lái)解決科學(xué)問(wèn)題。有時(shí)同一信息可以用來(lái)解答兩個(gè)或更多的問(wèn)題。

　　PART C

　　Direction: Every question or incomplete statement below is followed by five suggested answers or completions. Choose the one that is best in each case and then fill in the corresponding oval on answer sheet.

　　說(shuō)明：下面的每個(gè)問(wèn)題或不完整表述都帶有五個(gè)備選答案。選出最佳答案并填寫(xiě)在答題紙的相應(yīng)位置上。

　　EXAMPLE：

　　If the molar mass of NH 3 is 17g/mol, what is the density of this compound at STP?

　　1. 0.25g/L　　2. 0.76g/L　　3. 1.25g/L　　4. 3.04g/L　　5. 9.11g/L

　　The solution of this quantitative problem depends on the application of several principles. One principle is that the molar mass of a gas expressed in grams/mole will occupy 22.4L at standard temperature and pressure (STP). The other is that the density of a gas at STP is the mass of 1L of the gas. Therefore, 17g of ammonia (NH 3) will occupy 22..4L, and 1L is equal to 17g/22.4L or 0.76g/L. The correct answer is (B).

　　如果(NH 3)的分子質(zhì)量是17g/mol, 在標(biāo)準(zhǔn)狀況下這種化合物的密度是多少?

　　1. 0.25g/L　　2. 0.76g/L　　3. 1.25g/L　　4. 3.04g/L　　5. 9.11g/L

　　這個(gè)定量問(wèn)題的解決需要運(yùn)用幾個(gè)原理。其中之一是在標(biāo)準(zhǔn)狀況下(STP)一摩爾氣體所占的體積是22.4升。另一原理是在STP氣體的密度為1L氣體的質(zhì)量。因此，1mol氨氣的體積為22.4升，1L質(zhì)量就是17/22.4 g或者0.76 g。正確答案為(B)。