2024年5月29日 · The ionic compound SnBr4 is called tin (IV) bromide. It consists of tin cations with a +4 charge and bromide anions with a -1 charge.

2024年6月3日 · What is the ionic compound SnBr4 called? The ionic compound SnBr4 is called tin (IV) bromide. It consists of tin cations with a +4 charge and bromide anions with a -1 charge.

SnBr4(aq)+2Na2S(aq) 4NaBr(aq)+SnS2(s) Suppose a student adds 49.5 mL of a 0.448 M0 solution of SnBr4 to 39.7 mL of a 0.196 MM solution of Na2S. Identify the limiting reactant. Calculate the theoretical yield of SnS2. The student recovers 0.447 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.

The name of the compound SnBr4 is Select one: a. tin bromide b. tin (II) bromide c. tin tetrabromide d. tin (IV) bromide Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on.

Question: In the redox reaction 2MnBr3+ SnBr2 -----> 2MnBr2 + SnBr4 Mn undergoes an oxidation number change of six units MnBr3 is the substance oxidized SnBr2 is the oxidizing agent Sn undergoes an oxidation number change of two units

2022年12月4日 · The cation for SnBr4 is Sn^4+, which is the tin(IV) cation. What is the compound for tin and bromine? These compounds are tin bromides: SnBr2, SnBr4. Is Cl- a cation or anion?

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction. SnBr4(aq)+2Na2S(aq) 4NaBr(aq)+SnS2(s) Suppose a student adds 36.0 mL of a 0.602 M solution of SnBr4 to 49.2 mL of a 0.128 M solution of Na2S. Identify the limiting reactant. SnBr4 NaBr Na2S SnS2 Calculate the theoretical yield of SnS2. Please help!

The ionic compound SnBr4 is named tin(IV) bromide. AnswerBot. ∙ 7mo ago. This answer is:

SnBr4(aq)+2Na2S(aq) 4NaBr(aq)+SnS2(s) Suppose a student adds 45.0 mL of a 0.605 M solution of SnBr4 to 53.8 mL of a 0.181 M solution of Na2S. Identify the limiting reactant. Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq) 4NaBr(aq)+SnS2(s) Suppose a student adds 45.1 mL of a 0.418 M solution of SnBr4 to 46.6 mL of a 0.133 M solution of Na2S. Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.