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3 years ago

Which om sled below is amplo dl an abelis laeler in an evünment?

Chemistry

2 answers:

Alla [95]3 years ago

7 0

Answer:

Explanation:

i got D

Marysya12 [62]3 years ago

5 0

It’s D!
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In which reaction is it possible for nitrogen to change into carbon?

sertanlavr [38]

C. ionization reaction...
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4 years ago

If you start with 13 mole H2O and 50 grams of O2 what is the percent yield of H202 if 100 grams of H202 is actually made?

dmitriy555 [2]

Answer:

94.1 %

Explanation:

We firstly determine the equation:

2H₂O + O₂ → 2H₂O₂

2 moles of water react to 1 mol of oxygen in order to produce 2 moles of oxygen peroxide.

We convert the mass of oxygen to moles:50 g . 1mol /32g = 1.56 mol

Certainly oxygen is the limiting reactant.

2 moles of water react to 1 mol of oxygen.

13 moles of water may react to 13/2 = 6.5 moles. (And we only have 1.56)

As we determine the limiting reactant we continue to the products:

1 mol of O₂ can produce 2 moles of H₂O₂

Then 1.56 moles of O₂ will produce (1.56 . 2) = 3.125 moles

We convert the moles to mass: 3.125 mol . 34 g/mol= 106.25 g

That's the 100% yield or it can be called theoretical yield.

Percent yield = (Yield produced / Theoretical yield) . 100

(100g / 106.25 g) . 100 = 94.1 %

3 0

3 years ago

The energy released by combustion of 1 g of a substance is called the ________ of the substance. the energy released by combusti

Anastaziya [24]

Enthalpy of the substance

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4 years ago

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How to obtain pure aluminium oxide from sodium aluminate. pls help.

cupoosta [38]

Answer:

By Bayer process.

Explanation:

In the Bayer process, bauxite ore is heated in a pressure vessel along with a sodium hydroxide solution (caustic soda) at a temperature of 150 to 200 °C. At these temperatures, the aluminium is dissolved as sodium aluminate (primarily [Al(OH)4]−) in an extraction process.

6 0

3 years ago

In the laboratory you dissolve 19.1 g of ammonium fluoride in a volumetric flask and add water to a total volume of 375 . mL. Wh

11111nata11111 [884]

Answer:

1.376 M

Explanation:

The following data were obtained from the question:

Mass of ammonium fluoride (NH₄F) = 19.1 g

Volume of solution = 375 mL

Molarity of ammonium fluoride (NH₄F) =?

Next, we shall convert 375 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

375 mL = 375 mL × 1 L / 1000 mL

375 mL = 0.375 L

Next, we shall determine the number of mole in 19.1 g of ammonium fluoride (NH₄F). This can be obtained as follow:

Mass of NH₄F = 19.1 g

Molar mass of NH₄F = 14 + (4×1) + 19

= 14 + 4 + 19

Molar mass of NH₄F = 37 g/mol

Mole of NH₄F =?

Mole = mass /Molar mass

Mole of NH₄F = 19.1 / 37

Mole of NH₄F = 0.516 mole

Finally, we shall determine the molarity of the solution. This can be obtained as follow:

Volume of solution = 0.375 L

Mole of NH₄F = 0.516 mole

Molarity of NH₄F =?

Molarity = mole /Volume

Molarity of NH₄F = 0.516 / 0.375

Molarity of NH₄F = 1.376 M

Therefore, the molarity of the ammonium fluoride (NH₄F) is 1.376 M

3 0

3 years ago