What is true of a molecule?

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

$\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}$

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

$\text{Moles of NO} = \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}$

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

$\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}$

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

$\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}$

7. Calculate the mass of NH₃ reacted

$\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}$

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0

3 years ago

Does the temperature of the solvent (water) affect the rate of dissolving?

gregori [183]

Yes. Heating up the solvent gives the molecules more kinetic energy. The more rapid motion means that the solvent molecules collide with the solute with greater frequency and the collisions occur with more force. Both factors increase the rate at which the solute dissolves.

3 0

3 years ago

Part V Write the (a) half reaction of oxidation, (b) half reaction of reduction, & (c) the balanced redox reaction for all o

Eduardwww [97]

Answer:

Explanation:

a. Oxidation : 2O + 4e^- ------> 2O^2-

b. Reduction: 2Sr - 4e- -------> Sr^2+

c. Balanced redox reaction

2Sr + O2 ------------> 2Sr O

Oxidation and reduction can be defined by various means, addition of oxygen, removal of hydrogen, removal of electrons. For this reaction, this definition is used, oxidation is the loss of electrons while reduction is the gaining of electrons.

In (a) oxidation half reaction, the valency of oxygen is zero and then moves into lossing two electrons resulting into -2 valency.

In (b) reduction half reaction, the valency of Sr is zero and gains electrons resulting into valency of 2.

In the overall redox reaction, Sr and O2 with valency of 0 each reacts together and form SrO with valency of 2 and -2 respectively, which gives 0 and then balances the equation.

8 0

3 years ago

I don't know how to do this can I get the answers plz it's due in 1 hour

katrin [286]

Okay, so even if I just gave you the answers, your teacher needs work on it too so it'll be easier/better if I just explain how to do it.
Basically, both sides need to have the same number of molecules. To do this, we make charts. This is the first side of number one:
Na - 1
Mg- 1
F - 2
The subscript gives F two molecules, and the other ones only each have one. This is the second side:
Na- 1
Mg- 1
F- 1
So they're not equal. To fix this, we add coefficients. These are numbers that are going to appear in the front of each compound/element and changes the number of molecules of the WHOLE compound/element. We need two F on the second side, so we'll put a coefficient of 2 in front of NaF. The new chart for the second side is this:
Na- 2
Mg- 1
F- 2
Now we've fixed the F, but now Na is off! So let's go to the first side again and see what we can do. We can put a 2 in front of the Na. The new chart is this:
Na- 2
Mg -1
F- 2
Now both sides are the same. The full new equation is:
2Na + MgF(sub2) = 2NaF + Mg
Basically, do this for all of them. Feel free to ask more questions.

4 0

3 years ago

Read 2 more answers

Consider the following four structures: What is the relationship of I and II? What is the relationship of I and III? What is the

daser333 [38]

Answer:

The different structures are shown in the attachment.

I and II - structural isomers

I and III - Structural isomers

I and IV - structural isomers

II and III - structural isomers

II and IV - structural isomers

III and IV - stereoisomers

Explanation:

The knowledge of Isomerism is tested here; there are two types of isomerism ; structural and stereoisomerism.

Structural Isomers have similar molecular and different double bond positioning, these occurs mostly in ALKENE FAMILY.

Stereo-isomers have the same molecular formular and similar patterns but differ in their spatial arrangement. trans and cis are typical examples of stereo-isomers.

From the question; Relationship between I and II is that they are structural isomers since they have the same molecular formula, but different bond atom arrangement and infact they are the same compound.

Relationship between I and III is that they are structural isomers with similar molecular formular but differ in the double bond position.

Relationship between I and IV is that they are structural isomers with similar molecular formula but different double bond arrangement.

Relationship between II and III is that they are structural isomers with similar molecular formular but different double bond position

Relationship between II and IV is that they are also structural isomers with the same molecular formular but different double bond position.

Relationship between III and IV is that they are stereo-isomers with same molecular formula but different spatial arrangement, hence cis and trans.

4 0

3 years ago