Saturday
March 28, 2015

Homework Help: Chemistry Who step by step

Posted by Ken on Sunday, May 11, 2014 at 6:29pm.

A 0.8870g sample of a mixture of nacl and nan03 is dissolved in water, and the solution is then treated with an excess of AGN03 to yield 1.913g of agcl. Calculate the percent by mass of the nacl in the mixture.

1.913g / 143 = 0.0133
x/58 ( 0.8870g -x) y/85 = 0.0133
0.01529 - 0.01724x + 0.01176x = 0.0133
x= 0.00199/0.00548
= 0.363/0.8870g x 100
=41%
100g - 41g = 59g

The right answer 87.87%

Who helps me step by step

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

Chenistry - A 0.8870g sample of a mixture of nacl and kcl is dissolved in water ...
chemistry Who helps step by step - 0.887g sample of a mixture of nacl and kcl is...
chemistry who helps me for me - A 0.887g sample of a mixture of nacl and kcl is ...
Chemistry Damon step bystep for me - A 0.887g sample of a mixture of nacl and ...
Gen. Chemistry - A 0.887 g sample of a mixture of NaCl and KCl is dissolved in ...
chemistry step by step for me - 0.6g mixture of licl and bal was treated with ...
chemistry - 1. An 887.0 mg sample of a mixture containing only NaCl and KCl is ...
Chemistry - A 0.3146-g sample of a mixture of NaCl(s) and KBr(s) was dissolved ...
Chemistry Damon step by step for me - you are given a solid that is a mixture of...
chemistry - x gm og Ag was dissolved in HNO3 and the solution was treated with ...

Members