️ cus | copper(ii) sulfide react with hno3 | nitric acid produce cu(no3)2 | copper(ii) nitrate + h2o | water + no | nitrogen monoxide + s | sulfur. Condition cho cus tác dụng với dung dịch axit hno3, khí không màu thoát ra 3 c u + 8 h n o 3 → 3 c u (n o 3 ) 2 + 2 n o + 4 h 2 o, copper atom loses two electrons to form c u (i i) ion.
In this process, it changes its oxidation state from 0 to + 2. Thus, copper is oxidized. During the reaction, nitrogen present in nitric acid changes its oxidation state from + 5 to + 2.
Thus, it is reduced. Hence, copper acts as. 4hno3(浓)+cu=cu(no3)2+2no2+2h2o 8hno3(稀)+3cu=3cu(no3)2+2no+4h2o (1)反应速率浓硝酸比稀硝酸快。 （2)铜与浓硝酸的物质的量之比是1:4. 铜与稀硝酸反应时，物质的量之比为3:8 （3）二氧化氮是红棕色，一氧化氮是无色。 （4）在铜与浓硝酸的反应中，每个铜原子失去.
(3cu + 8hn{o_3} to 3cu{(n{o_3})_2} + 2no + 4{h_2}o). Số phân tử đóng vai trò là chất oxi hóa là: 3cu + 8hno3 to 3cu(no3)2 +.
6．下列化学反应中. 硝酸只表现氧化性的是 ( ) a．3cu+8hno3(稀)＝3cu(no3)2+2no↑+4h2o b．cuo+2hno3 (稀)＝cu(no3) 2 +h2o c．c+4hno3 (浓)＝co2 ↑+4no2↑+2h2o d．3ag+4hno3 (稀)＝3 agno3 +no↑ +2h2o. 摘要：6．下列化学反应中. 硝酸只表现氧化性的是 ( ) a．3cu+8hno3(稀)＝3cu(no3)2+2no↑+4h2o. 3cu + 8hno3 → 3cu(no3 )2 + 4h2o + 2no.
← prev question next question →. Asked feb 27, 2019 in chemistry by anjal (77. 1k points) from the equation : 3cu + 8hno 3 → 3cu(no 3 ) 2 + 4h 2 o + 2no.
The mass of copper needed to react with 63 g of nitric acid. Experts are tested by chegg as specialists in their subject area. We review their content and use your feedback to keep the quality high.
3cu (s) + 8hno3 (aq) + 3cu (no3)2 (aq) + 2no (g) + 4h2o () what volume of 0. 325 m hno is required to completely react with 25. 08 of cu? What volume of 0. 125 m hno3 (a) is required to. 根据反应3cu+8hno 3 =3cu(no 3) 2 +2no↑+4h 2 o 回答下列问题： （1）氧化剂是 ，氧化产物是 。 （2）氧化剂与氧化产物的物质的量比是 。 （3）当有2 mol hno 3 被还原时，被氧化的物质的质量为 ，反应中转移的电子数为 ;
（4）有下列6种物质：k 2 so 4 、k 2 so 3 、i 2 、h 2 so 4 、kio 3 、h 2 o组成一个氧化还原反应. 3cu(s)+8hno3 (aq)→3cu(no3 )2 (aq) +2no(g)+4h2 o(l) tentukan perubahan bilangan oksidasi. Sd matematika bahasa indonesia ipa terpadu penjaskes ppkn ips terpadu seni agama bahasa daerah
For the equation 3cu + 8hno3 → 3cu(no3)2 + 2no + 4h2o, how many units of no3 are represented on the products side? A 2 b 3 c 6 d 8 2 see answers. 3cu(s) + 8hno3(aq) 3cu(no3)2(aq) + 4h2o(l) + 2no(g) in the equation above, copper is.
+ 8hno 3(aq) 3cu(no 3) 2(aq) + 4h 2 o (l) + 2no (g) in the equation above, copper is a. An electron acceptor b. An oxidizing agent d.
A reducing agent correct answer: On product side we have cu (no3)2. View the full answer.
What element is oxidized in the following chemical reaction? 3cu + 8hno3 → 3cu (no3)2 + 2no + 4h20 oo on oh cu i have no. 3cu(s)+8hno3=3cu(no3)2+2no+4h2o(1) how many nitric oxide, no are formed after mixing if we have 0. 25mol cu with 0. 7 mol hno3.
(ii) iodide and aqueous lead(ii) nitrate is represented by the balanced formula equation. Coi2(aq) + pb(no3)2(aq) → pbi2(s) + co(no3)2(aq) give the balanced ionic equation for the reaction. 3cu + 8hno3 = 3cu(no3)2 + 4h2o + 2no | chemical equation.
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Equations with cu as reactant. 2agno 3 + cu → 2ag + cu(no 3) 2 cu + 2h 2 so 4 → 2h 2 o + so 2 + cuso 4 cu + s → cus view all equations with cu as reactant A)moles de no formadas a partir de 1;8 moles de cu b)gramos de agua formados por 1;5 moles de hno3 c)masa de cu necesaria para preparar 180g de cu(no3)2 d)moles de hno3 necesarios para preparar 0;75g decu(no3)2
