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elixir [45]
2 years ago
5

The percentage of silver in a solid sample is determined gravimetrically by converting the silver to Ag+ (aq) and precipitating

it as silver chloride. Failure to do which of the following could cause errors in the analysis?I Account for the mass of the weighing paper when determining the mass of the sampleII Measure the temperature during the precipitation reactionIII Wash the precipitateIV Heat the AgCl precipitate to constant mass(A) I & II
(B) I & IV
(C) II & III
(D) I, III, & IV
Chemistry
2 answers:
Wittaler [7]2 years ago
8 0

Answer:

D

Explanation:

I - You should account the mass of the weighing paper to reduce it from the total mass at the end of the process, having only the mass of the Silver Chloride.

II - The precipitation of the silve chloride will occur independently of the temperature, because the Kps of this salt is very low (Ksp = [Ag+] .[Cl-]).

III - Washing the precipitate will secure the purity of the final product, it won't allow any other contaminant to be in your precipitante which could change your final mass.

IV - You should heat you AgCl precipitate so it will be dry, because of that the mass you will obtain is only the mass of the weighing paper and the silver chloride and nothing else.

nika2105 [10]2 years ago
5 0

Answer:(D) I, III, & IV

Explanation:

Firstly, the precipitate needs to be washed to remove water soluble impurities adsorbed on the surface of the precipitate which could affect its mass. Secondly, the precipitate must be heated to constant mass to drive off solvent molecules that adhered to the solid and could inflate the mass of the precipitate thereby introducing error to the analysis. The mass of the weighing paper must be accounted for because the mass read from the balance is the mass of the sample+ weighing paper.

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What is the mass of nickel(ii) nitrate (182.71 g/mol) dissolved in 25.0 ml of 0.100 m ni(no3)2 solution?
Studentka2010 [4]
The mass  of  ni(NO3)2  that  dissolved  in  25.0 ml  of 0.100m  ni(NO3)2  solution  is  calculated  as   follows

fin  the  number  of  moles   =  molarity   x  volume in  liters

=25  x0.100/ 1000= 2.5  x10^-3  moles
mass  =  mass  x  molar  mass
= 2.5  x10^-3 moles x  182.71 g/mol  = 0.457  grams
4 0
2 years ago
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Benzene, C6H6, reacts with oxygen, O2, to form CO2 and H2O. How much O2 is required for the complete combustion of 1.0 mol C6H6?
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Complete with factors to stabilize
C6H6 + 15/2 O2 forms 6 CO2 + 3 H2O
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Just Lemons Lemonade Recipe Equation:
zalisa [80]

Answer:

Explanation:

Hello!

<em>Complete text:</em>

<em>Honors Stoichiometry Activity WorksheetInstructions: </em>

<em>Activity Two: Just Lemons, Inc. Production</em>

<em>Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.</em>

<em>Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients. </em>

<em>Water 946.36 g </em>

<em>Sugar 196.86 g </em>

<em>Lemon Juice 193.37 g </em>

<em>Lemonade 2050.25g</em>

<em>Leftover Ingredients?</em>

<em>Just Lemons Lemonade Recipe Equation:</em>

<em>2 water + sugar + lemon juice = 4 lemonade</em>

<em>Mole conversion factors:</em>

<em>1 mole of water = 1 cup = 236.59 g</em>

<em>1 mole of sugar = 1 cup = 225 g</em>

<em>1 mole of lemon juice = 1 cup = 257.83 g</em>

<em>1 mole of lemonade = 1 cup = 719.42 g</em>

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water  + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g  = 4*(719.42)g

    473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

<u>Water:</u>

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

<u>Sugar:</u>

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

<u>Lemon juice:</u>

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

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