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Anarel [89]
3 years ago
7

What volume of oxygen gas is needed to react completely with 0.626 L of carbon monoxide gas, CO, to form gaseous carbon dioxide?

Assume all volume measurements are made at the same temperature and pressure.
Chemistry
1 answer:
faltersainse [42]3 years ago
4 0

Answer:

0.296L of O₂ are needed for the reaction

Explanation:

This is the reaction:

O₂  + 2CO  →  2CO₂

To determine volume, we must work with density. But firstly we have to know the mass.

Density = mass/volume

Ratio is 2:1

CO density = CO mass / CO volume

1.18×10⁻³g/mL = CO mass / 0.626 L

First we must convert 0.626L to mL, because the units of density

0.626L = 626 mL

1.18×10⁻³g/mL = CO mass / 626 mL

CO masa = 1.18×10⁻³g/mL . 626 mL → 0.741 g

Let's convert to moles, the mass of CO ( mass . molar mass)

0.741 g / 28 g/mol = 0.0264 moles

As ratio is 2:1, we would need the half of moles, so (0.0264 / 2) = 0.0132 moles are required for the reaction.

Let's convert the moles to mass (molar mass  . moles)

32 g/mol . 0.0132 mol = 0.423 g

Now, we can work with oxygen density

O₂ density = O₂ mass / O₂ volume

1.43×10⁻³g/mL = 0.423g / O₂ volume

O₂ volume =  0.423g / 1.43×10⁻³g/mL → 296 mL

296 mL / 1000 = 0.296L

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The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes, (1) NO(g)
AVprozaik [17]

Answer:

The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ

Explanation:

Given the following reactions and their standard enthalpy changes:

(1) NO(g) + NO₂(g) → N₂O₃(g) ΔH o rxn = −39.8 kJ

(2) NO(g) + NO₂(g) + O₂(g) → N₂O₅(g) ΔH o rxn = −112.5 kJ

(3) 2 NO₂(g) → N₂O₄(g) ΔH o rxn = −57.2 kJ

(4) 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ

(5) N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ

You need to get the heat of reaction from: N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g)

Hess's Law states: "The variation of Enthalpy in a chemical reaction will be the same if it occurs in a single stage or in several stages." That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when verified in a single stage.

This law is the one that will be used in this case. For that, through the intermediate steps, you must reach the final chemical reaction from which you want to obtain the heat of reaction.

Hess's law explains that enthalpy changes are additive. And it should be taken into account:

  • If the chemical equation is inverted, the symbol of ΔH is also reversed.
  • If the coefficients are multiplied, multiply ΔH by the same factor.
  • If the coefficients are divided, divide ΔH by the same divisor.

Taking into account the above, to obtain the chemical equation

N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g)  you must do the following:

  • Multiply equation (3) by 2

(3) 2*[2 NO₂(g) → N₂O₄(g) ] ΔH o rxn = −57.2 kJ*2

<em>4 NO₂(g) →  2 N₂O₄(g)  ΔH o rxn = −114.4 kJ</em>

  • Reverse equations (1) and (2)

(1) <em>N₂O₃(g)  → NO(g) + NO₂(g) ΔH o rxn = 39.8 kJ</em>

(2) <em>N₂O₅(g) →  NO(g) + NO₂(g) + O₂(g)  ΔH o rxn = 112.5 kJ</em>

Equations (4) and (5) are maintained as stated.

(4) <em>2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ </em>

(5) <em>N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ </em>

The sum of the adjusted equations should give the problem equation, adjusting by canceling the compounds that appear in the reagents and the products according to the quantity of each of them.

Finally the enthalpies add algebraically:

ΔH= -114.4 kJ + 39.8 kJ + 112.5 kJ -114.2 kJ + 54.1 kJ

ΔH= -22.2 kJ

<u><em>The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ</em></u>

8 0
3 years ago
Density is an example of a​
Readme [11.4K]
A rock is obviously more dense than a crumpled piece of paper of the same size. A styrofoam cup is less dense than a ceramic cup.


hope this helps if it do mark me as brainlest thank you !!
7 0
3 years ago
Consider the following molecules and the description of the bonding present in each: CH3CH2CH2CH3CH3CH2CH2CH3 (C−CC−C and C−HC−H
denis23 [38]

CH3CH2CH2CH3 < CH3CH2CHO < CH3CHOHCH3

Explanation:

Boiling point trend of Butane, Propan-1-ol and Propanal.

Butane is a member of the CnH2n+2 homologous series is an alkane. Alkanes have C-H and C-C bonds which have Van der waals dispersion forces which are temporary dipole-dipole forces (forces caused by the electron movement in a corner of the atom). This bond is weak but increases as the carbon chain/molecule increases.

In Propan-1-ol(Primaryalcohol), there is a hydrogen bond present in the -OH group. Hydrogen bond is caused by the attraction of hydrogen to a highly electronegative element like Cl-, O- etc. This bond is stronger than dispersion forces because of the relative energy required to break the hydrogen bond. Alcohols (CnH2n+1OH) also experience van der waals dispersion forces on its C-C chain and C-H so as the Carbon chain increases the boiling point increases in the homologous series.

Propanal which is an Aldehyde (Alkanal) with the general formula CnH2n+1CHO. This molecule has a C-O, C-C and C-H bonds only. If you notice, the Oxygen is not bonded to the Hydrogen so there is no hydrogen bond but the C-O bond has a permanent dipole-dipole force caused by the electronegativity of oxygen which is bonded to carbon. It also has van der waals dispersion forces caused by the C-C and C-H as the carbon chain increases down the homologous series. The permanent dipole-dipole forces are not as easy to break as van der waals forces.

In conclusion, the hydrogen bonds present in alcohols are stronger than the permanent dipole-dipole bonds in the aldehyde and the van der waals forces in alkanes (irrespective of the carbon chain in Butane). So Butane < Propanal < Propan-1-ol

3 0
3 years ago
What volume would 4.5 moles of helium gas occupy at STP?
zzz [600]

Answer:

100.8 dm3

Explanation:

mol= volume      

       _______

       molar volume  

Simply transpose the formula      

volume=mol x molar volume  

           4.5molx22.4 mol/dm3

=   100.8dm3

NB  STP=22.4 dm3  or 22400 cm3

7 0
3 years ago
Explain the effects that different frequencies of electromagnetic radiation have when absorbed by matter.
Neko [114]
In order to answer this question we might first want to think about what is electromagnetic radiation. In essence it’s light, just some of the wavelengths are too long or too short for us to see.

We can think about it as two oscillating sinusoidal (goes up and down) waves, one is electric, the other is magnetic.

Because we’re dealing in waves, that means we can calculate their frequency, wavelength, amplitude (brightness) and period.

To calculate it we can use E=hc/lambda
Where E = jewels of energy
h = Planck’s constant
c = speed of light
Lambda = wavelength

It doesn’t really matter for this question what those things mean, just note that it takes more energy to have a shorter wavelength, or less energy to have a longer wavelength.

So now we can answer the question. Light of a longer wavelength has less energy than that of a shorter wavelength. So, when long wavelengths are absorbed by matter (atoms) they will give those atoms less energy. So, either it will pass through the object entirely or it will make the atoms vibrate a little bit more than they already are and we call that thermal energy, or heat.

If high energy wavelengths are passing through matter then they will be giving those atoms a lot of energy, sometimes even ionizing the atoms.
Which, if you’re a living thing can be very bad for your cells.

I hope that helps.
8 0
4 years ago
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