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Montano1993 [528]
3 years ago
5

Which of the following mixture types can be filtered to remove solute? (1 point) suspensions only colloids only suspensions and

colloids suspensions and solutions
Chemistry
2 answers:
mixer [17]3 years ago
8 0
The correct answer is suspensions only. The suspension is a heterogeneous mixture that contains solid particles that are largely enough to undergo sedimentation. Usually, these particles are about one micrometer which makes these solute to be very easy to be free from their solvent and be filtered.
Murljashka [212]3 years ago
7 0

The answer to your question is

suspensions only

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M of CH2O= 12+2+16=30 g/mol
So 30g/mol = 3.5 mol
=> 16g/mol*3.5 divide by 30 = 1.86 mol
I think
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According to the Arrhenius definitions of acids and bases, choose the acids from the list of acids and bases. Check all that app
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Which sample contains particles in a rigid, fixed, geometric pattern?(1) CO2(aq) (3) H2O(ℓ)(2) HCl(g) (4) KCl(s)
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How much calcium oxide would be made by the thermal decomposition of 25 grams of calcium carbonate? caco3 -> cao co2
stealth61 [152]

14 grams of calcium oxide.

Thermal decomposition of 25 grams of calcium carbonate would result in the production of 14 grams of calcium oxide.

We know that,

CaCO₃ → CaO + CO₂

First, the following quantities react and are generated according to the stoichiometry of the reaction, which is the relationship between the amounts of reagents and products in a chemical reaction:

  • CaCO₃: 1 mole
  • CaO: 1 mole
  • CO₂: 1 mole

Being:

  • Ca: 40 g/mole
  • C: 12 g/mole
  • O: 16 g/mole

<h3>the chemicals taking part in the reaction have the following molar masses:</h3>
  • CaCO₃: 40 g/mole + 12 g/mole + 3*16 g/mole= 100 g/mole
  • CaO: 40 g/mole + 16 g/mole= 56 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole=  44 g/mole

The following mass amounts of the compounds involved in the reaction then react and are created, according to the reaction's stoichiometry:

  • CaCO₃: 1 mole* 100 g/mole= 100 g
  • CaO: 1 mole* 56 g/mole= 56 g
  • CO₂: 1 mole*  44 g/mole= 44 g

<h3>The rule of three can then be utilized: </h3>

How much calcium oxide will be produced if, according to the stoichiometry of the reaction, 100 grams of calcium carbonate CaCO3 result in 56 grams of calcium oxide CaO and 25 grams of CaCO3?

mass of calcium oxide = \frac{25 grams of CaCO3 + 56 grams CaO}{100 grams of CaCO3}

mass of calcium oxide= 14 grams

14 grams of calcium oxide would be produced by thermal decomposition of 25 grams of calcium carbonate.

To learn more about thermal decomposition visit:

brainly.com/question/14949019

#SPJ4

5 0
1 year ago
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