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sukhopar [10]
3 years ago
11

How many milliliters of 0.2560 M KCl solution will contain 20.00 g of KCl?

Chemistry
1 answer:
lutik1710 [3]3 years ago
3 0

Answer:

1048 mL

Explanation:

Step 1: Given data

  • Concentration of the solution: 0.2560 M
  • Mass of KCl (solute): 20.00 g

Step 2: Calculate the moles corresponding to 20.00 g of KCl

The molar mass of KCl is 74.55 g/mol.

20.00 g × 1 mol/74.55 g = 0.2683 mol

Step 3: Calculate the volume of the solution

Molarity is equal to moles of solute divided by liters of solution.

M = moles of solute / volume of solution

volume of solution = moles of solute / M

volume of solution = 0.2683 mol / (0.2560 mol/L)

volume of solution = 1.048 L = 1048 mL

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3 years ago
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten c
Darya [45]

Answer:

0.382g

Explanation:

Step 1: Write the reduction half-reaction

Al³⁺(aq) + 3 e⁻ ⇒ Al(s)

Step 2: Calculate the mass of Al produced when a current of 100. A passes through the cell for 41.0 s

We will use the following relationships.

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  • The molar mass of Al is 26.98 g/mol.

The mass of Al produced is:

41.0s \times \frac{100C}{s} \times \frac{1mole^{-} }{96486C} \times \frac{1molAl}{3mole^{-} } \times \frac{26.98gAl}{1molAl} = 0.382gAl

7 0
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8 0
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How to solve this: a small dog has consumed a dangerous amount of apap. the vet has ordered: rx acetylcysteine 150mg/kg/stat. yo
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Given:
Stock dose/concentration of 20% Acetylcysteine (200 mg/mL)
150 mg/kg dose of Acetylcysteine
Weight of the dog is 13.2 lb

First we must convert 13.2 lb to kg:
13.2 lb/(2.2kg/lb) = 6 kg

Then we must calculate the dose:
(150 mg/kg)(6kg) = 900 mg

Lastly, we must calculate the dose in liquid form to be administered:
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3 years ago
Find the volume of hydrogen gas formed when 1.5g of aluminum reacts with aq NaOH at 27 degrees Celcius.
Rudiy27

Answer:

how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?

(A) n=m/M,

n(Al)=5.4/27=0.2 moles

n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles

Number of oxygen atoms= n(O2)*Avogadro's number

=0.15*6.02*10^23=9.03*10^22 oxgyen atoms

(B)

n=m/M

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m=n*M

m(O2)=0.0166666*32=0.53333 grams

8 0
2 years ago
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