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3 years ago

What steps would you take in a lab to prepare 1000 mL of a 2 M NaCl solution

Chemistry

1 answer:

amm18123 years ago

7 0

Answer:

The answer to your question is below

Explanation:

Data

Volume = 1000 ml

Concentration = 2M

molecule = NaCl

Process

1.- Calculate the number of moles of NaCl

Molarity = moles/Volume

-Solve for volume

moles = Molarity x Volume (liters)

-Substitution

moles = 2 x 1

-Result

moles = 2

2.- Determine the molar mass of NaCl

NaCl = 23 + 35.5 = 58.5 g

3.- Calculate the mass of NaCl to prepare the solution

58.5 g ----------------- 1mol

x ----------------- 2 moles

x = (2 x 58.5) / 1

x = 117g

4.- Weight 117 g of NaCl, place them in a volumetric flask (1 l), and add enough water to prepare the solution.

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Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

3 years ago

What salt is produced in each of the following neutralization reactions?

Pavlova-9 [17]

1. Salt is KNO₃

This is a strong acid - strong base reaction. HNO₃ is the strong acid and KOH is the strong base. 

H⁺ in the HNO₃ and OH⁻ of the KOH pair up and make H₂O(l). 

NO₃⁻ and K⁺ pair up to make KNO₃ salt. 

When writing chemical formulas positive ion comes first and second is negative ion. The charges should be switched. Since positive ion has +1 and negative ion has -1 after the switching off charges the product should be KNO₃.

Balance equation is 
HNO₃(aq) + KOH(aq) → H₂O(l) + KNO₃(aq)

2. Salt is Ca(NO₃)₂

This is a strong acid - strong base reaction. HNO₃ is the strong acid and Ca(OH)₂ is the strong base. 

H⁺ in the HNO₃ and OH⁻ of the Ca(OH)₂ pair up and make H₂O(l). 

Ca²⁺ and NO₃⁻ pair up to make Ca(NO₃)₂ salt. 

Positive ion is Ca²⁺ which has +2 charge and negative ion is NO₃⁻ which has -1 charge. From switching the charges Ca²⁺ gets 1 while NO₃⁻ gets 2. Hence, the salt should be Ca(NO₃)₂.

Balanced equation is
2HNO₃(aq) + Ca(OH)₂(aq) → 2H₂O(l) + Ca(NO₃)₂(aq)

3. Salt is CaCl₂

This is a strong acid - strong base reaction. HCl is the strong acid and Ca(OH)₂ is the strong base. 

H⁺ in the HCl and OH⁻ of the Ca(OH)₂ pair up and make H₂O(l). 

Ca²⁺ and Cl⁻ pair up to make CaCl₂ salt. 

Positive ion is Ca²⁺ which has +2 charge and negative ion is Cl⁻ which has -1 charge. By switching the charges Ca²⁺ gets 1 while NO₃⁻ gets 2. Hence, the salt should be CaCl₂.

Balance equation is
2HCl(aq) + Ca(OH)₂(aq) → 2H₂O(l) + CaCl₂(aq)

4. Salt is KCl

This is a strong acid - strong base reaction. HCl is the strong acid and KOH is the strong base. 

H⁺ in the HCl and OH⁻ of the KOH pair up and make H₂O(l). 

K⁺ and Cl⁻ pair up to make KCl salt. 

Positive ion is K⁺ which has +1 charge and negative ion is Cl⁻ which has -1 charge. By switching the charges K⁺ gets 1 and Cl⁻ also gets 1. Hence, the salt should be KCl.

Balance equation is
HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq)

5 0

3 years ago

Read 2 more answers

Many enzymes are inhibited irreversibly by heavy metal ions such as Hg2+ , Cu2+ , or Ag+ , which can react with essential sulfhy

3241004551 [841]

Answer:

The minimum molecular weight of the enzyme is 29.82 g/mol

Explanation:

Step 1: Given data

The volume of the solution = 10 ml = 10*10^-3L

Molarity of the solution = 1.3 mg/ml

moles of AgNO3 added = 0.436 µmol = 0.436 * 10^-3 mmol

Step 2: Calculate the mass

Density = mass/ volume

1.3mg/mL = mass/ 10.0 mL

mass = 1.3mg/mL *10.0 mL = 13mg

Step 3: Calculate minimum molecular weight

Molecular weight = mass of the enzyme / number of moles

Molecular weight of the enzyme = 13mg/ 0.436 * 10^-3 mmol

Molecular weight = 29.82 g/mole

The minimum molecular weight of the enzyme is 29.82 g/mol

7 0

3 years ago

3. Indicate the type of reaction represented by each equation:

BartSMP [9]

3. Is double replacement

6 0

3 years ago

Given 6.98 x 10 4 power grams of iron, calculate the moles of iron present

KiRa [710]

Answer:

1249.88 mol.

Explanation:

∵ no. of moles of Fe = mass of Fe/atomic weight of Fe.

∴ no. of moles of Fe = (6.98 x 10⁴ g)/(55.845 g/mol) = 1249.88 mol.

6 0

2 years ago