To separate solid particles from a liquid, you would use:

A 29.05 gram sample of cobalt is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 40.88 g. Determ

deff fn [24]

The empirical formula of the oxide is Co₂O₃.

Step 1. Calculate the mass of oxygen

Your reaction is

Cobalt + oxygen ⟶ cobalt oxide

29.05 g + x g ⟶ 40.88 g

According to the Law of Conservation of Mass, the total mass of the reactants must equal the total mass of the products. Thus,

29.05 g + x g ⟶ 40.88 g

x = 40.88 – 29.05 = 11.83

Step 2. Calculate the moles of each element

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Co to O.

Moles of Co = 29.05 g Co × (1 mol Co /(58.93 g Co) = 0.492 96 mol Co

Moles of O = 11.83 g O × (1 mol O/16.00 g O) = 0.739 38 mol O

Step 3. Calculate the molar ratio of the elements

Divide each number by the smaller number of moles

Co:O = 0.429 26:0.739 38 = 1:1.4999

Step 4. Multiply each number by a factor that makes the ratio close to whole numbers

Multiply by 2. Then

Co:O = 2:2.998 ≈ 2:3

Step 5: Write the empirical formula

EF = Co₂O₃

4 0

3 years ago

N=?

Mazyrski [523]

Okay so your answer for n is

7 0

3 years ago

PLZ HELP QUICKLY!!

frutty [35]

Answer:

Robert Hooke.

Anton van Leeuwenhoek.

Matthias Schleiden.

Theodor Schwann.

Rudolf Virchow.

Explanation:

6 0

3 years ago

A 10-gram sample of zinc loses 560 J of heat and has a final temperature of 100

grandymaker [24]

<h3>Answer:</h3>

Initial temperature is 243.59°C

<h3>Explanation:</h3>

The quantity of heat is calculated by multiplying the mass of a substance by its specific heat capacity and change in temperature.

That is; Q = m×c×ΔT

In this case;

Quantity of heat = 560 J

Mass of the Sample of Zinc = 10 g

Final temperature = 100°C

We are required to determine the initial temperature;

This can be done by replacing the known variables in the formula of finding quantity of heat,

Specific heat capacity, c, of Zinc = 0.39 J/g.°C

Therefore,

560 J = 10 g × 0.39 J/g°C × ΔT

ΔT = 560 J ÷ (3.9 J/°C)

= 143.59°C

But, since the sample of Zinc lost heat then the temperature change will have a negative value.

ΔT = -143.59°C

Then,

ΔT = T(final) - T(initial)

Therefore,

T(initial) = T(final) - ΔT

= 100°C - (-143.59°C)

= 243.59°C

Hence, the initial temperature of zinc sample is 243.59°C

5 0

3 years ago

The labeled images each represent the wave patterns found in the electromagnetic wave spectrum. Which image is correctly labeled

Nataly_w [17]

Https://us-static.z-dn.net/files/d15/c111c5e1b23135c61adec7b554629964.jpg

I believe the answer is the third option. Hope this helps! Please tell me if I am wrong or if there was an error in my answer... also sorry this answer is late.

6 0

3 years ago

Read 2 more answers