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3 years ago

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A chemist prepares a solution of barium chlorate by measuring out of barium chlorate into a volumetric flask and filling the fla

sk to the mark with water. Calculate the concentration in of the chemist's barium chlorate solution. Be sure your answer has the correct number of significant digits.

1 answer:

Elan Coil [88]3 years ago

6 0

Complete Question

A chemist prepares a solution of barium chlorate BaClO32 by measuring out 42.g of barium chlorate into a 500.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /molL of the chemist's barium chlorate solution. Be sure your answer has the correct number of significant digits.

Answer:

The concentration is $C = 0.28 \ mol/L$

Explanation:

From the question we told that

The mass of $Ba(ClO_{3})_2$ is $m_b = 42 \ g$

The volume of the solution $V_s = 500 mL = 500*10^{-3} L$

Now the number f moles of $Ba(ClO_{3})_2$ in the solution is mathematically represented as

$n = \frac{m_b}{Z_b}$

Where $Z_b$ is the molar mass of $Ba(ClO_{3})_2$ which a constant with a value

$Z_b = 304.23 \ g/mol$

Thus

$n = \frac{42}{304.23}$

$n = 0.14 \ mol$

The concentration of $Ba(ClO_{3})_2$ in the solution is mathematically evaluated as

$C = \frac{n}{V_2}$

substituting values

$C = \frac{0.14}{500*10^{-3}}$

$C = 0.28 \ mol/L$

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How would you make a 2.00L of 0. 500M sodium chloride solution. (assume you have a fully equipped lab with water); sketch, calc

Gwar [14]

Answer:

Sodium chloride solution:

First you need to calculate the mass of salt needed (done in the explanation), which is 58.44g. Then it have to be weighted in an analytical balance in a weighting boat and then transferred into a 2L volumetric flask that is going to be filled until the mark with distilled water.

Sulfuric acid dilution:

First you need to calculate the volume needed (done in the explanation), it is 16.6 mL. Using a graduated pipette one measures this volume and transfer it into a 2L volumetric flask that is already half filled with distilled water, and then one fills it until its mark.

Explanation:

Sodium chloride solution:

Each liter of a 0.500M solution has half mol, so 2L of said solution has 1 mol of salt. Sodium chloride molar mass is 58.44g/mol, so in 2L of solution there is 58.44g of salt. That`s the mass that`s going to be weighted and transferred to a 2L volumetric flask.

Sulfuric acid dilution:

This is the equation for dilution of solutions:

$c_{1} v_{1} =c_{2} v_{2}$

Where "c1" stands for the initial concentration (stock solution concentration), "v1" for the initial volume (volume of stock solution used), "c2" for the desired concentration and "v2" for the desired volume.

When we are diluting from a stock solution we want to know how much do we have to pipette from the stock solution into our volumetric flask. We do so by isolating the "v1" term from the dilution equation:

$v_{1} =\frac{c_{2} v_{2} }{c_{1} }$

in this case that would be:

$v_{1} =\frac{0.100 x2.0 }{12.0 }=0.0166L=16.6mL$

5 0

3 years ago

E. H2CrO4 + Sr(OH)2 →

katrin2010 [14]

Answer:

The answer to your question is SrCrO₄ + H₂O

Explanation:

Data

H₂CrO₄ + Sr(OH)₂ ⇒

We can notice that this is a Redox reaction or neutralization reaction because the reactants are an acid (H₂CrO₄) and a base (Sr(OH)₂). These reactions are also called double displacement reactions.

In these kind of reactions the products are always a binary or ternary salt and water.

Then, for this reaction,

H₂CrO₄ + Sr(OH)₂ ⇒ SrCrO₄ + H₂O

6 0

2 years ago

Read 2 more answers

a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what

jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$V$ - Volume, measured in liters.

$n$ - Molar quantity, measured in moles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

$\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}}$ (2)

Where:

$P_{1}$ , $P_{2}$ - Initial and final pressure, measured in kilopascals.

$V_{1}$ , $V_{2}$ - Initial and final volume, measured in liters.

$T_{1}$ , $T_{2}$ - Initial and final temperature, measured in Kelvin.

If we know that $P_{1} = 121\,kPa$ , $P_{2} = 202\,kPa$ , $V_{1} = 2.7\,L$ , $T_{1} = 288\,K$ and $T_{2} = 303\,K$ , the final volume of the gas is:

$V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}$

$V_{2} = 1.702\,L$

The gas will occupy a volume of 1.702 liters.

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3 years ago

the density of water at different temperature is listed in the table above. Based on this information we can predict that the de

son4ous [18]

look it nup on google should help

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3 years ago

1. (8pt) Using dimensional analysis convert 600.0 calories into kilojoules

Ivanshal [37]

Answer:

1. 2.510kJ

2. Q = 1.5 kJ

Explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, we can proceed as follows:

1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:

$1cal=4.184J\\\\1kJ=1000J$

Then, we perform the conversion as follows:

$600.0cal*\frac{4.184J}{1cal}*\frac{1kJ}{1000J}=2.510kJ$

2. Here, we use the general heat equation:

$Q=mC(T_2-T_1)$

And we plug in the given mass, specific heat and initial and final temperature to obtain:

$Q=236g*0.24\frac{J}{g\°C} (34.9\°C-8.5\°C)\\\\Q=1495.3J*\frac{1kJ}{1000J} \\\\Q=1.5kJ$

Regards!

7 0

2 years ago