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.

  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

Twitter(1) Twitter(2) Twitter(3) Twitter(Club Official) GitHub GitHub Site(WIP) Novels Telegram(Personal) Telegram(Channel) PlanetMC Modrinth Codeforces Wikidot

 title: hospital-diary author: VillagerVicky pages: #- Nov 26, 2023 晴 我生病了，被迫住进了医院。 我也说不清是什么问题，但我突然晕倒了，醒来后就在这里。 唉，来的第一天就觉得很无聊了。幸好我有手机，可以勉强解闷。 嘛，等出去了一定要好好逛一逛。这边风景很好，我还没怎么参观过。 #- Nov 27, 2023 晴 我看见了一群孩子，和我差不多年纪吧。 不知道为什么，我第一眼就注意到了她，似曾相识。 鼓起勇气，上去搭讪，他们很友好，我们成为了朋友。 这地方很小，没什么好逛的，我也不是来旅游的。 他们没什么玩的，却也很开心，真好啊。 医生说我的情况有些严重，看来一时半会走不了了。 #- Nov 28, 2023 雷雨 来到这里的第三天。 我无法再忍受了，且不提这里荒无人烟，而且晚上还有奇怪的怪物，周围除了电视塔什么都看不见...不是迫不得已谁想来医院呢？那些孩子们也没法缓解我的心情。 以及那黑暗的建筑是什么，窗户都没了... #- Dec 05, 2023 雨  果然，人在开心的时候是不会想写东西的。这么久过去了。 那些孩子很友好，但我还是和她玩的最多。虽然差不多大，但我们似乎更成熟一些，也更有话题。 有时看她治疗很痛苦，估计是很严重的病吧。我倒是还好，只是很虚弱，医生让我静养。我认为不过是没办法治疗吧。 #- 在对方最痛苦的时候，我们都互相陪伴着度过。这就是朋友的感觉吗？我很开心... 有些心疼，然而痛苦往往是不可避免的。 #- Dec 06, 2023 雪 下雪了，外面白茫茫的，要知道在这种地方雪并不常见。 她对我说，好想看看外面的世界是什么样子啊。 我很惊讶，问她，你难道没有出去过吗？或者说，你难道不是从外面来的吗？ 她说，自从自己记事以来，就是在医院里长大的，她听说自己是医生们捡来的孩子。 我不知道该如何安慰。 #- Dec 07, 2023 晴 我们又见面了，可是这次只有她一个人。 我问她，那些孩子们都到哪里去了，她说，都出去了，他们的病好了。 我说，好耶，等我们也好了，我们就一起出去玩，一起去找他们哦。 “好，好啊...可是，我出不去的，我活不了多久了。” “哈哈，那么我也活不了多久了，没关系的。” 希望你只是开玩笑。 #- Dec 10, 2023 阴 这几天，我们一起探...

Oxalate  V(V) forms a surprising complex: cis-[VO2(C2O4)2]3- whose ammonium salt (NH4)3[VO2(C2O4)2].2H2O is bright yellow and forms large, maybe even single crystals, but solution tends to be unsaturated till the end. Preparation of this is quite hard to describe: Mixing same molar NH4VO3, H2C2O4 and (NH4)2C2O4 together in solution is OK, but then it darkens soon even if you are very precise. You may use a few milliliter H2O2 to oxidize it back it turns red, and then in a few days it darkens and repeat H2O2 addition... finally it does not become black but brown, then go to vaporize at RT. Seems to be stable both as solid and solution, but wet reduced dark blue parts of the solid turns into red insoluble substance in a few months. If you managed to get some good seeds and saturated solution, you may grow very large crystals as usual. Solubility is so high that visible growth is observed every few hours, and extremely large crystals can be obtained in only a few days. Crystals t...

 An electrolysis cell is a container with two electrodes in it. In order to be effective we need partitioning: carefully hit a test tube at ground to remove its bottom, and add glass wool in it. Here we do reduction so anode is inside, cathode is outside. V2+ is very reducing so use paraffin liquid above solution. In a short period, about a few days, you get purple solution which is the end. Heat this solution with excessive ammonium sulfate and magnesium sulfate (V:(NH4)2:Mg=1:5:4)then cool, then you get purple crystals. Similar procedure however, can't dye alums due to unknown reasons. Probably, different ion radius or wrong temperature? (Cooling also failed and solution freezing is driving me mad) to be continued...

 Alums are compounds with the formula AB(SO4)2.12H2O with A=+1 alkali/ammonium and B=+3 hexaaquo ion. The larger A is, or the smaller B is, the better stability crystal has. Most members are not that stable as Tutton's salts and sometimes only solid solutions and cesium alums are stable. Cubic system, octahedral crystals. Three structures exist, alpha for most, beta for cesium alums except Group 9 ones, gamma for sodium alums. They are honest in colors, as aqueous solutions of such a highly charged ion usually undergoes hydrolysis. Selenate can replace sulfate, and such alums are highly soluble and are exclusively alpha. Known examples: Al: Colorless. Best known, with K and NH4 common and stable. Na also possible but very unstable. Sc: Colorless. Unstable. Ti: Purple? Highly reducing both in solid and solution. V: Blue or green or what? Reducing. My procedure gave nothing but alum with solution. Better make pure compound. Cr: Dark purple, very famous. K mixing with KAl is the most ...

 Welcome back! Now, let's first dissolve some CoSO4 or MnSO4 into 5M sulfuric acid, together with ammonium alum. To do so, add 0.02 mol of ammonium sulfate and aluminium sulfate to 100mL of acid, then 0.02 mol of the transition metal one. Now heat to dissolve these all, then cool down for a night to let excess alum crystallize. They are just alum of course. From now on different strategies are needed for them. Mn: Mn3+ is stable when H+ is very concentrated and Mn2+ is much more concentrated than Mn3+, so we should electrolysis without partitioning, and only a little part of Mn2+ is oxidized. Finally solution looks like this: Finely red. Cool this solution down to 0 degree celcius or so gives orange crystals, and throwing alum into initial solution gives it a cover. All crystals MUST be dried carefully! Solution can be used again of course. It seems that crystals are unstable when [Mn3+] is too high. This is hard to control. Reduction is simple though, just carefully add oxalic aci...

 Vanadium is the 23rd element, lies in period 4 and group VB. Being an early transition metal it is somewhat similar to main group elements, especially P, but can also be reduced to the +3 and +2 states, common in transition metals. Its +4 state mainly exist as [VO]2+ which is a stable terminal oxo complex. It is somewhat expensive and toxic, so use carefully. Before exploring more, I wanna make V3+ and V2+ as double salts. I obtained this element as sodium metavanadate (NaVO3), which makes it a little troublesome to remove unwanted Na+. If you can get NH4VO3 and you want an ammonium salt product, just add acid and reduce it. NaVO3 is a fine white powder that dissolves slowly in water. Heat to dissolve it. Then add sulfuric acid carefully until pH becomes 2~3. Now the solution is dark yellow and much red stuff precipitated. The solid is V2O5, and is very sticky. Filtering is quite hard and sick, so just pour off the liquid carefully, add water, then pour again, to wash it. You may ...

A2B(SO4)2.6H2O where A is univalent, K or larger, normally NH4, and B is divalent. B must exist as hexaaquo ion. So, there is in fact no water of crystallization. These compounds are usually highly stable due to crosslinking hydrogen bonds, being resistant to oxidation and erosion in air even if B is Cr, Fe or V. Simply mix the sulfates in solution, and they quickly reacts to form double salt crystals. Even solid sulfates are OK, as a saturated solution of Tutton's salt is not saturated for the simple salts. Monoclinic. If you use no fishing string like me, you may get two different shapes due to different directions(see below), and you can manually turn the diamond one to the brick one by simply turning the seed upside down(?). It seems that +3 cation strongly impacts transparency. Note: ratios below are that of mass of saturated solution at ambient conditions, NOT solid as it would be very hard to prepare or recycle. This doesn't matter in most cases, but if you wanna use mor...

About the ions Although +3 oxidation state exists for all transition metals between Sc and Co, they are almost always unfriendly. While the common Cr3+ is stable and Fe3+ is mildly oxidizing and highly acidic, the closely related Mn3+ and Co3+ are exceedingly unstable at normal ranges. Instead of directly telling you the phenomena I'd like to show thermodynamic data first. E0(Mn3+/Mn2+)=+1.49V E0(MnO2/Mn3+)=+0.95V E0(Co3+/Co2+)=+1.92V E0(O2/H2O)=+1.229V The first two lines may need an explanation: if we make a battery of the first minus the second, it turns out that Mn3+ can release much electricity(energy) when it becomes Mn2+ and MnO2, as the net potential is highly positive, about +0.5V. That is to say, Mn3+ is very prone to disproportionation. The third line could just be called horrible: Co3+ is much higher than O2, so oxidizing water is quite easy! In fact this is just at pH=0 or [H+]=1mol/L, and things become much worse at neutral pH as H+ lies on the right side of the equat...