In the field of electrochemical biosensing based on DNA probes immobilised at electrode surfaces the [Ru(NH3)6]3+ redox marker is commonly used for quantitative characterisation and/or detection purposes. Although it is generally assumed that the interactions between the hexaammineruthenium(III) ...
设在[RuBr2(NH3)4]+中Ru的化合价为x,x-1×2+0×4=+1,x=+3;含有空轨道的金属阳离子为中心离子,所以中心离子为Ru3+、有孤对电子的原子、分子或离子为配体,所以配体为Br-、NH3,配位数就是配体的个数,所以配位数为2+4=6,故选D。解题步骤 氧气的三种化学性质包括:1.氧气是一种氧化剂,它能够...
Ru(NH3)6Cl3 元素 质量百分比氯106.359g氯106.359g釕101.07g釕101.07g氮84.0402g氮84.0402g氫18.1429g氫18.1429gRu(NH3)6Cl3 原子#氫18氫18氮6氮6氯3氯3釕1釕1 🛠️ 计算摩尔质量 指令 这个程序决定一种物质的分子量。输入物质的分子式,它将计算出元素的组合物和质量的化合物中的每个元素的总质量。
故答案为:(c^2(N_2)•c^6(H_2O))/(c^4(NH_3)•c^3(O_2));③由图可知,520℃时氮气的物质的量为0.2mol,NO物质的量为0.2mol,根据氮原子守恒,则Δn(NH3)=0.2mol×2+0.2mol=0.6mol,氨气转化率=(△n(NH_3))/(n_(起始)(NH_3))×100%=(0.6mol)/(1mol)×100%=60%,...
Hexaammineruthenium(III) trichloride forms a Z = 12 monoclinic crystal, which is related to a simple cubic f.c.c. structure by, dominantly, a rotation of a third of the hexaammineruthenium(III) fragments. The cell contains four crystallographically inequivalent Ru(NH3)(6) fragments, but two...
The junction is formed by bringing into contact two mercury-drop electrodes whose surfaces are covered by COO−-terminated self-assembled monolayers (SAMs) and immersed in a basic aqueous solution of Ru(NH3)6Cl3. The electrical behavior of the junction, which is contacted at its edges by ...
百度试题 题目6.下列配离子中未成对电子数是多少?估计其磁矩各为多少(BM) (1)[Ru(NH3)3]2+(低自旋状态);(2)[Fe(CN)6]3-(低自旋状态) (3)[Ni(H2O)6]2+; (4)[Ven)3]3 5) [Cocl相关知识点: 试题来源: 解析反馈 收藏
Finally, add together the total mass of each element to get the molar mass of Ru(NH3)6Cl3: 101.07 g/mol + 84.0402 g/mol + 18.14292 g/mol + 106.359 g/mol =309.61212 g/mol 5. Find Mole Fraction To find the mole fraction and percentage of each element in Ru(NH3)6Cl3, divide each...
A.根据分析,阳极电极反应为4OH--4e-=O2↑+2H2O,阴极电极反应为:O2+2H++2e-=H2O2,电解池的总反应为:2H2O+O2=2H2O2,双氧水处理氨水,反应为:3H2O2+2NH3=N2↑+6H2O,两式合并,则该过程的总反应为:4NH3+3O2=2N2+6H2O,故A错误; B.利用电解法制H2O2,在该电解池中,Ir-Ru惰性电极有吸附O2作用为氧气得...
A.根据分析,阳极电极反应为4OH--4e-=O2↑+2H2O,阴极电极反应为:O2+2H++2e-=H2O2,电解池的总反应为:2H2O+O2=2H2O2,双氧水处理氨水,反应为:3H2O2+2NH3= N2↑+6H2O,两式合并,则该过程的总反应为:4NH3+3O2=2N2+6H2O,故A错误; B.利用电解法制H2O2,在该电解池中,Ir-Ru惰性电极有吸附O2作用为氧气得...