4. This is the idea that energy can be transformed from one form to another, but it

Four grams of magnesium react with an excess of dilute hydrochloric acid. Calculate the volume in liters of hydrogen gas collect

Helga [31]

Sorry I am confuse 3

5 0

3 years ago

Rank the following fertilizers in decreasing order of mass percentage of nitrogen:

charle [14.2K]

<h3>Answer:</h3>

NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

<h3>Soution:</h3>

In (NH₄)₂HPO₄:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄ = 132.06 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age = 28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age = 21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄ = 132.14 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age = 28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age = 21.18 %

In KNO₃:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of KNO₃ = 101.10 g.mol⁻¹

Mass %age = Mass of N / M.Mass of KNO₃ × 100

Mass %age = 14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age = 13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄ = 115.03 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age = 14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age = 12.17 %

In NH₃:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of NH₃ = 132.14 g.mol⁻¹

Mass %age = Mass of N / M.Mass of NH₃ × 100

Mass %age = 14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age = 82.20 %

In NH₄NO₃:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of NH₄NO₃ = 80.04 g.mol⁻¹

Mass %age = Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age = 28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

Mass %age = 34.98 %

5 0

3 years ago

If 2.0 mol of Zn is mixed with 3.0 mol<br> of HCl, which reactant will be limiting?

Soloha48 [4]

The following Balanced Reaction will take place:

Zn + 2HCl → ZnCl₂ + H₂

In the question, we have 2 moles of Zinc and 3 moles of HCl for this reaction

<u>Amount of HCl required to completely react with 2 moles of Zn:</u>

Since we need 2 moles of HCl for every mole of Zn, we will need 2(2) = 4 moles of HCl for every 2 moles of Zn

<u>Identifying the Limiting Reagent:</u>

But we are only given 3 moles of HCl where we need 4 moles to completely react.

So, since HCl is in less amount, it is the Limiting Reagent

8 0

2 years ago

Acetylene gas (C2H2) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide

Luda [366]

A. The balanced equation for the reaction is

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

B. The volume of the oxygen gas required to burn 0.700 L of acetylene gas is 1.75 L

C. The volume of carbon dioxide gas produced is 1.4 L

D. The volume of water vapor produced is 0.7 L

<h3>A. Balanced equation </h3>

The balanced equation for the reaction between acetylene gas (C₂H₂) and oxygen gas (O₂) is given below

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

<h3>B. How to determine the volume of oxygen </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

Since the reaction occurred at standard temperature and pressure, we can thus say that:

From the balanced equation above,

2 L of C₂H₂ reacted with 5 L O₂.

Therefore,

0.7 L of C₂H₂ will react with = (0.7 × 5) / 2 = 1.75 L of O₂

Thus, 1.75 L of O₂ is needed for the reaction

<h3>C. How to determine the volume of carbon dioxide </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 4 L of CO₂

Therefore,

0.7 L of C₂H₂ will react to produce = (0.7 × 4) / 2 = 1.4 L of CO₂

Thus, 1.4 L of CO₂ were produced

<h3>D. How to determine the volume of water. </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 2 L of H₂O

Therefore,

0.7 L of C₂H₂ will also react to produce 0.7 L of H₂O

Thus, 0.7 L of H₂O was produced

Learn more about stoichiometry:

brainly.com/question/14735801

8 0

2 years ago

If a car is traveling 100 km/h and comes to a stop in 3 minutes,what is acceleration of a passenger who is using vehicle restrai

Rama09 [41]

We are given –

Final velocity of car is, v= 0
Initial velocity of car is, u= 100 km/hr
Time taken, t is = 3 minutes or 180 sec

Here–

$\qquad$ $\pink{\bf \longrightarrow Initial\: velocity = 100 \:km/hr}$

$\qquad$ $\sf \longrightarrow Initial\: velocity = \dfrac{ 100 \times 1000}{3600} \:m/s$

$\qquad$ $\pink{ \bf \longrightarrow Initial\: velocity = 27.78\: m/s}$

Now –

$\qquad$ ____________________________

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{Final\: Velocity -Initial \:Velocity }{Time}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = \dfrac{v -u}{t}}$

$\qquad$ $\sf \longrightarrow Acceleration = \dfrac{(0- 27.78)}{1800}$

$\qquad$ $\sf \longrightarrow Acceleration =\cancel{ \dfrac{- 27.78}{1800}}$

$\qquad$ $\purple{\bf \longrightarrow Acceleration = -0.15 \: m/s^2}$

$\qquad$ _______________________________

6 0

2 years ago