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3 years ago

What is 4,678,679 dL in scientific notation

1 answer:

8 0

4.7e^6 is how I would write it. You simply count in from the right side where the decimal is until you reach the point just before the 4.

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WILL MARK BRANILEST

ahrayia [7]

Answer:

FeCl3 is the limiting reactant

O2 is in excess

Theoretical yield Cl2 = 9.84 grams

The % yield is 96.5 %

Explanation:

Step 1: Data given

Mass of FeCl3 = 15.0 grams

Moles O2 = 4.0 moles

Mass of Cl2 produced = 9.5 grams

Step 2: The balanced equation

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

Step 3: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 15.0 grams / 162.2 g/mol

Moles FeCl3 = 0.0925 moles

Step 4: Calculate limiting reactant

FeCl3 is the limiting reactant. Because we have way more (more than ratio 3:4) moles O2 than FeCl3. It will completely be consumed (0.0925 moles). O2 is in excess. There will react = 0.069375 moles O2

There will remain 4.0 - 0.069375 = 3.930625 moles

Step 5: Calculate moles Cl2

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

For 0.0925 moles FeCl3 moles we'll have 6/4 * 0.0925 = 0.13875 moles Cl2.

Step 6: Calculate mass Cl2

Mass Cl2 = moles * molar mass

Mass Cl2 = 0.13875 moles * 70.9 g/mol

Mass Cl2 = 9.84 grams

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (9.5 grams / 9.84 grams ) * 100%

% yield = 96.5 %

The % yield is 96.5 %

4 0

3 years ago

A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut

Irina-Kira [14]

Answer:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

It is a stichiometry problem.

We should write the balance equation of the mentioned chemical reaction:

2Al + 3CuCl₂ → 3Cu + 2AlCl₃.

It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.

Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

n = mass / molar mass

The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.

The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.

From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.

∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with (2:3) ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.

So, the answer is:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

5 0

4 years ago

What are the benefits of a lilac bonnet?

Tatiana [17]

Answer:

Maintains natural hair moisture.

Prevents hair frizz.

Reduces hair breakage & tangles.

Longer lasting hair styles.

Explanation:

3 0

3 years ago

When you have finished adding the necessary amount of NaOH to your beaker, what color will the litmus paper turn?

Svetllana [295]

Answer: The red litmus paper turns blue on dipping in NaOH solution.

Explanation:

Litmus paper is the indicator that detects the nature of the solution, whether it is acidic or basic.

There are 2 types of litmus paper:

Red litmus paper: This paper will turn blue if it is dipped in basic solution and will remain as such if it is dipped in acidic solution.
Blue litmus paper: This paper will turn red if it is dipped in acidic solution and will remain as such if it is dipped in basic solution.

NaOH is a strong base, so when a red litmus paper is dipped in the beaker having necessary amount of NaOH, the red litmus paper turns into blue.

7 0

3 years ago

Determine the molality of an aqueous solution that is 16.5 percent urea by mass

yarga [219]

Take a hypothetical sample of exactly 100 grams of the solution.

(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea

((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O

(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea

7 0

3 years ago