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3 years ago

Que unidad utilizamos para medir un átomo de oxígeno??

Chemistry

2 answers:

Leya [2.2K]3 years ago

5 0

Answer:

Los químicos eligen definir la AMU como 1/16 de la masa promedio de un átomo de oxígeno que se encuentra en la naturaleza; es decir, el promedio de las masas de los isótopos conocidos, ponderado por su abundancia natural. Los físicos, por otro lado, lo definieron como 1/16 de la masa de un átomo del isótopo oxígeno-16 (16O).

Aleks [24]3 years ago

4 0

Answer:

Im pretty sure you use mass

Explanation:

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Which beaker has the least thermal energy?

Debora [2.8K]

Answer:

Breaker A

Explanation:

Because the temperature is cooler in A then the rest

3 0

3 years ago

Read 2 more answers

Scenes A and B depict changes in matter at the atomic scale:

kaheart [24]

Scene B depicts chemical change in matter at atomic change.

Composition distinguishes a chemical reaction from a physical reaction. In a chemical process, the makeup of the components changes; in a physical change, the appearance, smell, or straightforward exhibition of a sample of matter changes without changing its composition. Despite the fact that we refer to them as physical "reactions," nothing is actually changing. A change in the substance in question's elemental composition is necessary for a reaction to occur. Therefore, from now on, we will simply refer to bodily "reactions" as physical changes.

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brainly.com/question/23693316

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5 0

1 year ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

3 years ago

An 8 oz. bottle of energy drink contains 6.0 g of protein, 2.0 g of fat, and 16.3 g of carbohydrate. The fuel value of this ener

valina [46]

Answer: The correct option is d) 460 kJ

Explanation:

We are given:

Content of fat in energy drink = 2.0 g

Content of protein in energy drink = 6.0 g

Content of carbohydrate in energy drink = 16.3 g

Also,

The fuel value of fat = 38 kJ/g

The fuel value of protein = 17 kJ/g

The fuel value of carbohydrate = 17 kJ/g

So, the fuel value of the energy drink will be:

Total fuel value = $(2.0g\times 38 kJ/g)+(6.0g\times 17 kJ/g)+(16.3g\times 17 kJ/g)$

Total fuel value = $[76+102+277]=460kJ$

Hence, the correct option is d) 460 kJ

4 0

3 years ago

Four liquids are described in the table below. Use the second column of the table to explain the order of their freezing points,

KIM [24]

Transcribed image text: Four liquids are described in the table below. Use the second column of the table to explain the order of their freezing points, and the third column to explain the order of their boiling points. For example, select '1' in the second column next to the liquid with the lowest freezing point. Select '2' in the second column next to the liquid with the next higher freezing point, and so on. In the third column, select '1' next to the liquid with the lowest boiling point, '2' next to the liquid with the next higher boiling point, and so on. Note: the density of water is 1.00g/mL .

7 0

2 years ago