. Using improved chemistry equipment in the late 1700s, chemists

4 Nitrogen monoxide reacts with oxygen like this:

olga2289 [7]

a. mol O₂=0.5

b. volume O₂ = 25 cm³

c. i. the total volume of the two reactants = 75 cm³

c. ii. the volume of nitrogen dioxide formed = 50 cm³

<h3>Further explanation</h3>

Reaction

2NO(gas) + O₂(gas) ⇒ 2NO₂ (gas)

a.

mol NO = 1

From the equation, mol ratio NO : O₂ = 2 : 1, so mol O₂ :

$\tt \dfrac{1}{2}\times 1=0.5$

b.

From Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles

Because mol ratio NO : O₂ = 2 : 1, so volume O₂ :

$\tt \dfrac{1}{2}\times 50~cm^3=25~cm^3$

c.

i. total volume of reactants : 25 cm³+ 50 cm³=75 cm³

ii. the volume of nitrogen dioxide formed :

mol ratio NO : NO₂ = 2 : 2, so volume NO₂ = volume NO = 50 cm³

6 0

2 years ago

Mrs. Rushing fills a balloon with hydrogen gas to demonstrate its ability to burn. Which combination could she use to produce th

ale4655 [162]

Answer:

Mg and HCI

Explanation:

3 0

3 years ago

Read 2 more answers

What is the net ionic equation for 2AgNO3 + 2NaOH = Ag2O + 2NaNO3 + H2O

soldi70 [24.7K]

Answer:

2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)

Explanation:

Step 1: RxN

2AgNO₃ + 2NaOH → Ag₂O + 2NaNO₃ + H₂O

Step 2: Define states of matter

2AgNO₃ (aq) + 2NaOH (aq) → Ag₂O (s) + 2NaNO₃ (aq) + H₂O (l)

Step 3: Total Ionic Equation

2Ag⁺ (aq) + 2NO₃⁻ (aq) + 2Na⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + 2Na⁺ (aq) + 2NO₃⁻ (aq) + H₂O (l)

Step 4: Cancel out spectator ions

2Ag⁺ (aq) + 2OH⁻ (aq) → Ag₂O (s) + H₂O (l)

7 0

3 years ago

The density of lead is 11.3g/cm^3 and the density of chromium is 7.14g/cm^3. If we have one gram of each metal, what is the rati

Lady bird [3.3K]

The ratio of the volume of lead to the volume of chromium is 22 : 35

Density is simply defined as the mass of a substance per unit volume of the substance.

<h3>Density = mass / volume </h3>

To solve the question given above, we'll begin by calculating the volume of lead and chromium. This can be obtained as follow:

Mass of lead = 1 g

Density of lead = 11.3 g/cm³

<h3>Volume of lead =? </h3>

Density = mass /volume

11.3 = 1 / Volume

Cross multiply

11.3 × Volume = 1

Divide both side by 11.3

Volume = 1 / 11.3

<h3>Volume of lead = 0.088 cm³</h3>

<h3>For Chromium:</h3>

Mass of chromium = 1 g

Density of chromium = 7.14 g/cm³

<h3>Volume of chromium =? </h3>

Density = mass /volume

7.14 = 1 / Volume

Cross multiply

7.14 × Volume = 1

Divide both side by 7.14

Volume = 1 / 7.14

<h3>Volume of chromium = 0.14 cm³</h3>

Finally, we shall determine the ratio of the volume of lead to the volume chromium. This can be obtained as follow:

Volume of lead = 0.088 cm³

Volume of chromium = 0.14 cm³

<h3>Ratio of the volume of lead to chromium =? </h3>

Ratio = Volume of lead / Volume of chromium

Ratio = 0.088 / 0.14

Ratio = 22 / 35

<h3>Ratio = 22 : 35</h3>

Therefore, the ratio of the volume of lead to the volume chromium is 22 : 35

Learn more: brainly.com/question/17262276

5 0

3 years ago

A solution is made using 148.5 g of dimethyl ether (MM = 46.07 g/mol) and 90.0 g of methanol (MM = 32.04 g/mol). What is the mol

Mariulka [41]

Answer:

0.534

Explanation:

Mole fraction can be calculated using the formula:

Mole fraction = number of moles of solute ÷ number of moles of solvent and solute (solution).

In this question, solute is dimethyl ether while the solvent is methanol.

Mole (n) = mass (M) ÷ molar mass (MM)

Mole of solute (dimethyl ether) = 148.5 ÷ 46.07

= 3.22moles.

Mole of solvent (methanol) = 90 ÷ 32.04

= 2.81moles.

Total moles of solute and solvent = 3.22 + 2.81 = 6.03moles.

Mole fraction of dimethyl ether = number of moles of dimethyl ether ÷ number of moles of solution (dimethyl ether + methanol)

Mole fraction = 3.22/6.03

= 0.534

6 0

3 years ago