Questions LLC
Login
or
Sign Up
Ask a New Question
Science
Chemistry
a solution of an unknown metal chloride (0.08870 g ) was treated with silver nitrate to yield AgCl ( 1.9132 g ). calculate the percentage by mass of chlorine in the unknown chloride
1 answer
I worked this for you below.
You can
ask a new question
or
answer this question
.
Similar Questions
a solution of an unknown metal chloride (0.08870 g ) was treated with silver nitrate to yield AgCl ( 1.9132 g ). calculate the
Top answer:
I worked this for you below.
Read more.
A 0.6753 g sample of an unknown metal was
converted to the nitrate, MCl2, then a solution of the nitrate was treated with silver
Top answer:
The story you tell of the metal is not very good; somehow the nitrate and chloride became mixed up.
Read more.
a solution of an unknown metal chloride (0.8870 g) was treated with silver nitrate to yield (1.9132 g). calculate the percentage
Top answer:
You obtained 1.9132 g AgCl. Convert that to grams Cl. The easy way to do that is to multiply by a
Read more.
a solution of an unknown metal chloride (0.8870 g) was treated with silver nitrate to yield (1.9132 g). calculate the percentage
Top answer:
See your post above.
Read more.
If 48.9g of barium chloride dihydrate, BaCl2.2H20, was treated with excess silver nitrate solution. (show formula)
a). how many
Top answer:
Here is a worked example of a stoichiometry problem. Just follow the steps.
Read more.
An aqueous solution of silver nitrate contains 5.00 g of silver nitrate. To this solution is added sufficient barium chloride to
Top answer:
You can do this one of several ways, first, as a reaction to precipate the silver. The second, is
Read more.
An anhydrous metal chloride, MCln, (1.500 g) is dissolved in water and the solution reacts with silver nitrate, requiring
Top answer:
(0.4734 dm^3)(0.5000 mol/dm^3) = ____moles Ag+ Moles Cl- = moles Ag+ Grams of Cl- = (35.453
Read more.
a portion (10.0 ml)of a solution containing the chloride anion was treated with silver nitrate to yield a precipitate (0.1713 g)
Top answer:
Convert 0.1713 g AgCl to g Cl moles AgCl = 0.1713/molar mass AgCl moles Cl = moles Ag since there is
Read more.
A 21.5 g sample of nickel was treated with excess silver nitrate solution to produce silver metal and nickel(II) nitrate. The
Top answer:
The balanced chemical equation is as follows: Ni(s) + 2AgNO3(aq) ==> Ni(NO3)2(aq) + 2Ag(s) Let X =
Read more.
If a .750g sample of magnesium chloride is dissolved in water and then treated with a silver nitrate solution how many grams of
Top answer:
0.750g of MgCl2*(1 mol/95.211g)=0.007877 mol of MgCl2 *****1 mol of MgCl2=2 moles of AgCl This is
Read more.
Related Questions
One method for reclaiming silver metal from silver chloride results in a 94.6%
yield. Calculate the actual mass of silver that
An excess of silver nitrate solution was added to 10.0cm3of sodium chloride solution, and 0.717g of silver chloride was
A technician mixes a solution containing silver nitrate with a solution containing sodium chromate. 2.89 g of precipitate is
A 500 mL potassium chloride solution was prepared by dissolving
potassium chloride in distilled water. 25.0 mL of the solution
a 6.54 g sample consisting of a mixture of silver nitrate and sodium nitrate is dissolved in water. this mixture then reacts
What mass of silver chloride, AgCl will precipitate from a solution that contains 1.50 g of, calcium chloride, CaCl if an excess
A total charge of 96.5 kC is passed through an electrolytic cell. What mass of silver metal will be produced from a silver
problem 2.
a sample of calcium chloride weighs 1.64g. it is dissolved in water and silver nitrate solution added.a precipitate of
According to the following reaction, how many grams of copper(II) chloride are necessary to form 0.984 moles silver chloride?
sil
what mass of silver(1)nitrate is needed to prepare 100cm3 of silver(1)nitrate solution,concentration 0.2mol/dm3?
the mass of one