How many significant figures does 0.040040 have ?

Describe an experiment for the preparation and collection of Oxygen sodium peroxide

vitfil [10]

Answer:

Making oxygen

Oxygen can be made from hydrogen peroxide, which decomposes slowly to form water and oxygen:

hydrogen peroxide → water + oxygen

2H2O2(aq) → 2H2O(l) + O2(g)

The rate of reaction can be increased using a catalyst, manganese(IV) oxide. When manganese(IV) oxide is added to hydrogen peroxide, bubbles of oxygen are given off.

Apparatus arranged to measure the volume of gas in a reaction. Reaction mixture is in a flask and gas travels out through a pipe in the top and down into a trough of water. It then bubbles up through a beehive shelf into an upturned glass jar filled with water. The gas collects at the top of the jar, forcing water out into the trough below.

To make oxygen in the laboratory, hydrogen peroxide is poured into a conical flask containing some manganese(IV) oxide. The gas produced is collected in an upside-down gas jar filled with water. As the oxygen collects in the top of the gas jar, it pushes the water out.

Instead of the gas jar and water bath, a gas syringe could be used to collect the oxygen.

5 0

3 years ago

_______ is the application of scientific knowledge to solve problems or create new products.

astraxan [27]

Having resources? i guess...
maybe it is critical thinking?

3 0

3 years ago

Read 2 more answers

Are these right? U guys r the best on this app!!

Pepsi [2]

Yes, they are right.Very good hand writing! You did a very great job on showing your work.

8 0

3 years ago

Read 2 more answers

Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . Wha

Dmitry [639]

Answer:

7.5 g

Explanation:

There is some info missing. I think this is the original question.

<em>Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Write the balanced equation

H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄

Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid

The molar mass of phosphoric acid is 98.00 g/mol.

$4.9 g \times \frac{1mol}{98.00g} = 0.050mol$

Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid

The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.

Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate

The molar mass of ammonium phosphate is 149.09 g/mol.

$0.050mol \times \frac{149.09 g}{mol} = 7.5 g$

6 0

3 years ago

A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc

Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

<em>1 mole of aniline reacts per mole of HCl</em>

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

7 0

3 years ago