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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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The following unbalanced equation illustrates the overall reaction by which the body utilizes glucose to produce energy: C6H12O6
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Answer:

the conversion factor is f= 6  mol of glucose/ mol of CO2

Explanation:

First we need to balance the equation:

C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)

the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:

f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction

f = 6 moles of CO2 / 1 mol of glucose = 6  mol of glucose/ mol of CO2

f = 6 mol of CO2/ mol of glucose

for example, for 2 moles of glucose the number of moles of CO2 produced are

n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2

3 0
3 years ago
Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

Therefore;

L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

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Answer:

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Explanation:

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