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Paul [167]
3 years ago
8

Describe the preparation of 2.000 L of 0.0500 M AgNO3 (169.87 g/mol) from the primary-standard-grade solid.

Chemistry
1 answer:
Tanzania [10]3 years ago
5 0

Answer:

Transfer 1.6987 grams of AgNO₃ into a mixing container and add solvent up to but not to exceed 2.000 Liters total volume. Mix and apply as needed.

Explanation:

A simple formula for determining the amount of reagent needed for a specified solution from a manufacturer's 'solid' reagent** is ...

grams solute needed = Molarity x Volume of Solution in Liters x Molecular Weight of solute

Determination of mass of solute needed:

∴ grams AgNO₃ needed* = M·V·mol. wt = (0.0500M)(2.000L)(169.87g·mol⁻¹) = 1.6987 grams AgNO₃

*assumes 100% pure reagent.

Preparation:

Transfer 1.6987 grams of AgNO₃ into a mixing container and add solvent up to but not to exceed 2.000 Liters total volume. Mix and apply as needed.

________________________________

NOTE => FYI => If a manufacturer's liquid form reagent is used, one must divide calculated grams by density of liquid supplied and gives volume in milliliters (ml) needed to prepare a specified solution.

Vol reagent needed for soln prep in milliliters = (Molarity·Volume in Liters·formula wt of reagent)/(density of stock liquid concentrate)

That is,  Vol(ml) = [(M)(V)(f.wt.)] / density of stock

Prep: Transfer needed volume (ml) of liquid stock into mixing vessel and add solvent up to but not to exceed total volume of solution needed.

<em>NOTE => Caution when applying to strong acids. 1st add a small quantity of water such that water + volume of acid is less than total volume needed followed by strong acid, then dilute to the needed total volume. Such prevents solution from overheating because of high rate of ionization and flashing out of container.</em>    

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Answer:

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A big interest that you have that leds you to pick a job related to that interest.

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Answer:

M HCl sln = 12.0785 M

Explanation:

  • molarity (M) [=] mol/L
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∴ mass sln = 100.0 g

∴ δ sln = 1.19 g/mL

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∴ molar mass HCl = 36.46 g/mol

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2 years ago
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Show the mechanism for the following reaction conducted at –5 °C in CCl4: cyclohexene bromine yields a dibromocyclohexane Draw w
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Answer:

See attachment.

Explanation:

In the first step, a cyclic structure with a positive bromine is formed. The bromine shares the positive charge with the two carbons that it is bonded to, so the carbons are partially positive.

The second bromine atom then attacks the carbon center, coming in from below the first bromine atom ("backside attack") where the antibonding orbital of the second bromine atom is.

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To determine molecular formula, we first need to find out its empirical formula,
Carbon. Hydrogen. Nitrogen. Oxygen
Mass. 49.98g. 5.19g. 28.85g. 16.48g
Mole. 4.165. 5.19. 2.06. 1.03
Divide 4. 5. 2. 1
by
smallest
So by comparing the mole ratio from the table above, i hope u understand the table
The empirical formula is C4H5N2O
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(C4H5N2O) n= 194.19
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