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3 years ago

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Suppose you needed a 0.0250 M sodium thiosulfate solution to conduct 6 titrations. Explain how you would make up this solution u

sing a 0.4782 M stock solution of sodium thiosulfate. Include all necessary measured quantities, lab techniques and the glassware you would use.

1 answer:

AysviL [449]3 years ago

4 0

Explanation:

According to the law of dilution,

$M_{1}V_{1} = M_{2}V_{2}$

The given data is as follows.

$M_{1}$ = 0.4782, $V_{1}$ = ?

$M_{2}$ = 0.025 , $V_{2}$ = 250 mL

Hence, we will calculate the value of $V_{1}$ as follows.

$V_{1} = \frac{M_{2} \times V_{2}}{M_{1}}$

= $\frac{0.025 \times 250}{0.4782}$

= 13.07

Thus, we can conclude that we need 13.07 mL 0.4782 M sodium thiosulfate solution using pipette.

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The boiling points for the homonuclear diatomic molecules in the halogen family are 85 K, 238 K, 332 K, 457 K and 610 K. Which s

patriot [66]

Answer:

$Cl_{2}$ has boiling point of 238 K

Explanation:

Boiling point depends on different intermolecular force such as molecular wight, dipole-dipole attraction force, hydrogen bonding, ionic attraction force.

Homonuclear diatomic molecules are covalent non-polar molecules and thereby free from dipole-dipole attraction force, hydrogen bonding and ionic interaction forces.

Hence, boiling point of homonuclear diatomic molecules depends solely on molecular weight.

We know, higher the molecular weight of a molecule, higher will be its boiling point. This phenomenon can be realized in terms of increasing london dispersion force with increase in molecular weight.

Decreasing order of molecular weight of halogen molecules :

$I_{2}$ > $Br_{2}$ > $Cl_{2}$ > $F_{2}$

So, decresing order of boiling point of halogen molecules:

$I_{2}$ > $Br_{2}$ > $Cl_{2}$ > $F_{2}$

Hence $Cl_{2}$ has boiling point of 238 K

4 0

3 years ago

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

2 years ago

What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC

Mkey [24]

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

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3 years ago

The pH at the equivalence point will be a. less than 7.00 b. equal to 7.00 Allohalineup of a sto quivalence point of a titration

True [87]

Answer: c. greater than 7.00

Explanation: The equivalence point of a titration is when all the base is consumed by the acid. When a strong base and a strong acid react, the medium is neutralized because is produced water and salt (which won't suffer hydrolysis). How water's pH is 7, in this type of titration the pH of the equivalence point will be at pH=7. But on titration of a weak acid with a strong base, the reaction of the equivalence point produces water and the conjugate base of the acid. Because the acid is weak, their conjugate base will be strong and will suffer hydrolysis, producing hydroxyl ions, elevating the pH of the water and making it greater than 7.

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2 years ago

Which particle makes the atom an unstable isotope?

Galina-37 [17]

Instability of an atoms nucleus can result from an excess of either neutrons or protons . So neutrons and protons .

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2 years ago