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Tatiana [17]
3 years ago
8

A chemist must dilute of aqueous potassium dichromate solution until the concentration falls to . He'll do this by adding distil

led water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Phantasy [73]3 years ago
5 0

The question is incomplete; the complete question is;

A chemist must dilute 99.4 mL of 152 mM aqueous potassium dichromate (K_2Cr_2O_7) solution until the concentration falls to 55.0 mM He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits.

Answer:

0.275 L

Explanation:

From C1V1 = C2V2

Where;

C1= initial concentration of the solution 152 × 10^-3 M

V1= initial volume of the solution = 99.4 × 10^-3 L

C2 = concentration after dilution = 55 × 10^-3 M

V2 = volume after dilution = the unknown

V2 = C1 V1/C2

V2 = 152 × 10^-3 × 99.4 × 10^-3 / 55 × 10^-3

V2 = 0.275 L

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2Al + 6HCl → 2AlCl3 + 3H2<br> If 85.0 grams of HCl react, how many moles of H2 are produced?
Murrr4er [49]

Answer:

1.17 mol

Explanation:

Step 1: Write the balanced equation

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

Step 2: Calculate the moles corresponding to 85.0 g of HCl

The molar mass of HCl is 36.46 g/mol.

85.0 g × 1 mol/36.46 g = 2.33 mol

Step 3: Calculate the number of moles of H₂ produced from 2.33 moles of HCl

The molar ratio of HCl to H₂ is 6:3.

2.33 mol HCl × 3 mol H₂/6 mol H₂ = 1.17 mol H₂

8 0
2 years ago
A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
Verdich [7]

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
What is the value of for this aqueous reaction at 298 k? a b↽−−⇀c dδ°=20. 46 kj/mol
sineoko [7]

The value for this aqueous reaction at 298 k? a b↽−−⇀c dδ°=20. 46 KJ/mol is 9.91 mol. in equilibrium

<h3>What is an aqueous reaction in equilibrium?</h3>

When a chemical reaction happens at the liquid state and the formation of reactant and product is the same then the reaction is known as an aqueous reaction in equilibrium denoted by K.

δG = − R T ln

          R = universal gas constant 8.313

          δG= 20. 46 kj/mol

           T =  298 k or 24.4 in celcius.

substituting the value in the equation.

20. 46 kj/mol = 8.313 × 24.4 in celcius × K

K =  8.313 × 24.4 in celcius / 20. 46 kj/mo

k = 9.91 mol .

Therefore, The value of this aqueous reaction at 298 k? a b↽−−⇀c dδ°=20. 46 KJ/mol is 9.91 mol. in equilibrium

Learn more about the aqueous reaction in  equilibrium, here:

brainly.com/question/8983893

#SPJ4

5 0
2 years ago
Which energy is found in the nucleus of an atom?
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It is called nuclear energy.
6 0
2 years ago
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Are insert gases used in fluorescent​
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Hydrofluorocarbons (HFCs), perfluorocarbons (PFCs), sulfur hexafluoride (SF6) and nitrogen trifluoride (NF3).
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