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Alona [7]
1 year ago
7

A student forgets to weigh a mixture of sodium bromide dihydrate and magnesium bromide hexahydrate. Upon strong heating, the sam

ple loses 252.1 mg of water. The mixture of anhydrous salts reacts with excess AgNO₃ solution to form 6.00X10⁻³ mol of solid AgBr. Find the mass % of each compound in the original mixture.
Chemistry
1 answer:
stira [4]1 year ago
4 0

The mass percent of each compound is

  • (NaBr*2H2O)=85.71 percent
  • (MgBr2*6H2O)=14.29 percent

<h3>The mass percent equation is as follows.</h3>

The formula mass percent is the most effective way to express mass percent: mass percent = (mass of chemical x total mass of compound) times 100. Add 100 to the top value to get the value expressed as a percentage.

<h3>What percentage by mass does a solution have?</h3>

By multiply the grams of solute per gram of solution by 100, it is possible to get the mass percent of a solution.

The molar mass of each element and the mass of each element contained in a mole of the compound are both solved for in the Mass Percent formula.

Learn more about Mass Percent here  

brainly.com/question/26150306

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Its molecules are made up of 60 carbon atoms joined together by strong covalent bonds. Molecules of C 60 are spherical. There are weak intermolecular forces between molecules of buckminsterfullerene. These need little energy to overcome, so buckminsterfullerene is slippery and has a low melting point.

Explanation:

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3 years ago
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__________ have an electric charge of -1.60 × 10-19 C (-e).
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For the reaction 2kclo3(s)→2kcl(s)+3o2(g) calculate how many grams of oxygen form when each quantity of reactant completely reac
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First, we need to get the molar mass of:

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From the given equation we can see that:

every 2 moles of KClO3 gives 3 moles of O2

when mass = moles * molar mass

∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g

and the mass of O2 then = 3 mol * 32g/mol = 96 g

so, 245.2 g of KClO3 gives 96 g of O2

A) 2.72 g of KClO3: 

when 245.2 KClO3 gives → 96 g  O2

   2.72 g KClO3 gives →  X

X = 2.72 g KClO3 * 96 g O2/245.2 KClO3

    = 1.06 g of O2

B) 0.361 g KClO3:

when 245.2 g KClO3 gives → 96 g O2

     0.361 g KClO3 gives → X

∴ X = 0.361g KClO3 * 96 g / 245.2 g

       = 0.141 g of O2

C) 83.6 Kg KClO3:

when 245.2 g KClO3 gives → 96 g O2

       83.6 Kg KClO3 gives  →  X

∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3

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D) 22.4 mg of KClO3:

when 245.2 g KClO3 gives → 96 g O2

        22.4 mg KClO3 gives → X

∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3

      = 8.8 mg of O2

     

 


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