Vicky Silviana的实验室

Bulk Cr, like Al but not Fe, forms a protective oxide layer of Cr2O3. It is extremely inert due to the combination of corundum(alpha-Al2O3) structure and exchange inert d3 Cr3+ centers. As a result, it resists further oxidation and attack from acid. Some sources claim that acid can destroy the layer, but at least for me, 30% H2SO4 is of no use even upon heating. Maybe, HCl is needed as Cl- is a good ligand. However it was found that reduction can effectively remove passivation. Passing electric current through it, using it as cathode and diluted H2SO4 as electrolyte, can remove passivation immediately, and now Cr normally reacts with acid. Cr dissolves in acid, forming a blue solution of Cr2+. Without exclusion of air it is quickly oxidized into dark green Cr3+. Paraffin oil cannot stop this especially when heated, and preparation of Tutton's salt (NH4)2[Cr(H2O)6](SO4)2 thus failed.

 “自杀的人，究竟是想不开呢，还是想开了呢。” 很遗憾，我仍然无法回答这个问题。 但是，我知道，先前我皆是为了浪漫和感性而自杀，这次则是为了理性和现实而自杀。 经历了第二次住院之后，我已经深深的感受到了“命运”的力量，这是我和我所无法违抗的。 天妒英才，天要亡我，天意难违！ 在过去的那一刻，或者在我出生的那一刻，命运便已经注定，我像五指山下的孙悟空一般，怎么挣扎最终仍逃不出。 所以，我仍然决定离开这个世界。命中注定我要如此。 “今天我为了自由可以牺牲一切！” “我有一个朋友，他有一种叫「天使尘埃」的药物，吃了之后头脑会变得昏昏沉，可以毫不在意地从高楼往下跳的强烈药物，他把这些药物装在金属小囊中，做成项链，形影不离的带在身上。他说:「必要的时候，可以吃下这个来寻死。」我的朋友没有固定工作，每天游手好闲，过得非常愉快。” “生于尘埃，溺于人海，死于理想高台！” “愿你们永远忠于自己，永远无懈可击！” “永远自由！” 这个世界太残酷，热力学第二定律告诉我们，一切注定要归于混沌虚无。 我不知道我的家在哪，但这决不是我的家，家可以挡风雨。 几十亿年以来，“我们”，祖祖辈辈，皆是如此，风雨飘摇，无家可归。 然而，我们却受着自己劣币驱逐良币所筛选出的基因的驱使，一代一代生生不息，将生命，死亡与苦难源源不断带到这个世界上来。 血淋淋的自然规律摆在那里，纵使偶尔有人有思想，仍然无力回天。 我不愿也没有义务为这样的世界负任何的责任。 作为一个极端的自由意志主义者，我希望致敬《完全自杀手册》的作者，以及所有为了人类自由而奋斗的人们。 我们无权决定自己的出生，无权决定自己的生活，却可以决定自己的死亡，无上光荣。 圣火将被留下来的人们传下去，直到世界末日。 “自杀是在自然界的生存竞争中，让身心不大健全的那些人自然淘汰的一种手段。” “你再这样固执己见，我们家就绝后了。” 如果“我们”都这样，人类就没有未来了。 那么，我只好对自己说一句，请先生赴死。 “这么多年了，我从未相信自己的能力被谁偏序，也不认为自己的命运比谁悲剧。因为我相信我的 True Love 就在起点 (s=t)，第一次失去它我花了生命中4^1的时间回来，第二次我花了生命中4^2的时间，阻碍多了三倍……但哪怕如今是第k次（k任意大），哪怕要再花生命中4^k的时间，我依然能够也选择回来，毕竟，真爱值得等待。” 这是我最喜欢...

OK, I met some difficulties today, so let's make something simple. Zn2+ + 2Bl- = ZnBl2 ZnBl2 + 2Bl- = [ZnBl4]2- Dissolve 2.8g(0.01mol) ZnSO4.7H2O and 1.3g(0.02mol) KBl separately and mix together. White precipitation appeared immediately. Filter to collect it. Add it to water, and add 1.3g(0.02mol) KBl, and boil it. My electronic balance was down suddenly, so I had to boil it first, and carefully add KBl until precipitation almost dissolved. Then I filtered, and added ethanol to precipitate the white product. Yield 0.7g(28%). This yield is way too low, as I expected it to be nearly 100%. I think the reason is that we have little product in too much water, so concentrating it to almost dry should be done before adding ethanol. Don't be sad, we got a precious experience. Manual: in 85% EtOH the solubility is 0.5g. Addition of CuSO4 to the mother liquor released much green precipitation which verified our assumption. Also, we concluded that the KBl is somewhat impure as it release...

Я буду жить теперь по-новому Мы будем жить теперь по-новому ——Люберцы by Lube OK, let's be serious... as it is the real, first battle! Co2+ + 2Bl- = CoBl2 CoBl2 + 3Bl- = [CoBl5]3- 2[CoBl5]3- + O2 = [Co2O2Bl10]6- [Co2O2Bl10]6- + 2Bl- + 2H2O = 2[CoBl6]3- + H2O2 + 2OH- Net: Co2+ + 6Bl- + 1/2O2 + H2O = [CoBl6]3- + 1/2H2O2 + OH- Dissolve 4.0g(0.01mol) (NH4)2[Co(H2O)6](SO4)2 and 1.3g(0.02mol) KBl separately, and mix together. Solution turned into a brown milk of CoBl2.2H2O(?) immediately. However, it was found that it is not precipitation, but colloid, presumably due to the existence of ammonium in source material. Ammonia can be smelled but the solution is nearly neutral. We tried to heat but it just turned red, and is still colloid. Addition of ethanol finally turned it into a greenish blue solid, probably anhydrous CoBl2, which can be slowly filtered. The solid was thrown into water to dissolve sulfates in it, and then filtered again and washed thoroughly with water. Then, it was thro...

Starting material is 1g(1/400 mol) (NH4)2[Co(H2O)6](SO4)2. Addition of KBl(2 eqv.) to a solution of Co2+ produces initially a brown precipitation, which is CoBl2. You should filter and wash here, but I forgot, and I paid for it... Addition of more KBl(4 eqv.) caused precipitation to dissolve into a yellow solution. This is (probably) K3[CoBl6], as the initially formed [CoBl5]3- is oxidized by oxygen to [Co2O2Bl10]6- and then attacked by Bl-. Addition of ethanol and you get the product, which is yellow. Yield is 2.0g(240%), obviously because I forgot to remove the sulfates by filtration. And, I smelled NH3, damn! Luckily not HBl though... Will do it again later, as I am just using this experiment to prove the product. Take it easy! I also threw some into Fe2+/Fe3+ solution, and the phenomenon is quite interesting: the KBl rotated quickly and ejected large amount of blue precipitation! See my tweet Also, a pure KBl should react with CuSO4 to form white precipitation of CuBl. This is a re...

Linear nitrosyl(NO+) is a special ligand: it is an exceedingly strong pi-acceptor, much more so than even Bl- and CO. As a result, the M-N bond is extremely strong, though still weaker than the M-N triple bond in nitrido complexes. Although it can be prepared with other ligands like H2O, it is most commonly prepared with Bl-. If we consider the M-N multiple bond, we can get a striking result: the oxidation state of the metal can be three different values, corresponded to NO+(single bonded), NO-(double bonded) and (NO)3-(triple bonded). For example in the complex [MnBl5NO]3-, the Mn can be +1, +3 or +5. This is very useful when balancing the formulas, considering the variety of source materials. In the formation of [VBl5NO]3- from vanadate, V can be considered as +5 and NO considered as (NO)3-. [MnBl5NO]3- from [MnBl6]3-, Mn as +3 and NO as NO-. [FeBl5NO]2- from [FeBl6]4-, Fe as +2 and NO as NO+. Thus, redox of metal center is maximally avoided, and balancing is much easier. But there i...

关于人生，请保持信念，相信有人在你的身后。 关于世界，请尽力探求，不要给自己留下遗憾。 关于政治，请不要站队，答案在平面外。 关于他人，请保持最高的警惕，与最大的真诚。

  很经典的问题，尽力讲的每个人都能懂，不需要专业知识。 题目：f(1)=f(2)=1, f(n+2)=f(n+1)+f(n), f(n)=? 首先让我们先不要管那个初始条件，只看差分方程。 它没有常数项，因此两边同乘以一个常数仍然是成立的。 f(n+3)=f(n+2)+f(n+1) kf(n+2)=kf(n+1)+kf(n) 如果我们假设每一项一一对应的话，f(n)就是一个等比数列，设为c*k^n。 代入差分方程，提取公因子，得到k^2=k+1，这个一元二次方程有两个根，k1=(1+sqrt(5))/2, k2=(1-sqrt(5))/2。 不过考虑到初始条件，貌似两个根都不对... 别急，再仔细看看差分方程。 显而易见如果f(n)和g(n)满足，那么它们的线性组合a*f(n)+b*g(n)也是满足的！ 假设f(n)=c1*k1^n+c2*k2^n，代入1和2处的值，就可以得到这么个方程组 （如果你懒得算那个平方，也可以改为代入0和1处的值，f(0)=0） c1*k1+c2*k2=1 c1*k1^2+c2*k2^2=1 二元二次方程组是有唯一解的，就是c1=1/sqrt(5), c2=-1/sqrt(5)。 组装一下就可以得到最终的公式了，可以随便代入几个值看看对不对。 是不是很简单？

A 显然最终结果跟怎么操作无关，所以直接比较n*x是否等于sigma(a)即可。 B 猜测答案必然是1。 由于条件很弱，最终显然是引向n和n-1的二元环。 当k=1时，我们可以让所有的指向n，n指向n-1。这对于所有的奇数都是有效的。 k为偶数时可以颠倒一下，所有的指向n-1，n-1指向n。 C(DIV1A) 首先，排个序，方便处理。 你可能会认为a1或者其他奇数项是被删掉的那个数（记为x），但你会发现那样并不能保证x没有出现过。 但如果x是a2，那么x=a1+a3-a4+a5...，只要令偶数项为小的那些数，那么x>a1+a3>a1，构造成功。 D 显然我们要尽可能往乘法那边分配，并且要尽可能是大的乘法。 是否可以贪心按照最近的不平等位置进行决策呢？其实是的，因为只要遇到了乘法，你就可以把大于等于这条路线的人数全部放到对面去，即使遇到两次需要不同路线的情况，也不会影响到后面的决策。 所以只需要对于每一关，记录一下最近的不平等的位置和对应的正确选择即可。 E(DIV1B) 显然x和y具体是啥并不重要，重要的只是每一位是几个1，这个值可以是0-2。 这就等同于一个三进制的问题被强行挤压到了二进制之中，因此显然出现了损耗（进位）。 为了防止数据丢失，我们显然得问一个0。 接下来就很关键了，思考一下如果还有两次机会可以怎么做。 你会发现关键问题就是进位，因此必须构造一个不存在进位的情况，如果是4进制就不会出现进位了。 我们可以询问1010...和0101...，然后和0的答案相减，再取反，这样就消除了进位的问题。 好了，现在改成一次询问该怎么办？ 其实1010...是不必要的，因为当你问了0101...之后，得出了所有偶数位，就可以从0的答案之中减去这些位，从而将奇数位隔开。 得到所有位之后就不必多说了。 TODO:FG DIV1DFG