All categories

2 years ago

Which type of fuel do you predict might have the

Chemistry

1 answer:

nadya68 [22]2 years ago

7 0

Answer:

If the choices were coal, petroleum, and natural gas, I would say that petroleum has the greatest impact on the CO₂ levels in the atmosphere. I chose petroleum because of transportation, such as cars, trucks, planes, and more.

You might be interested in

EX: The periodic table is arranged in order of increasing

Otrada [13]

Answer:

Valency electrons.

Explanation:

6 0

3 years ago

From the Henderson-Hasselbalch equation, explain how the ratio [Al/[HA] changes with changing pH

Kryger [21]

Answer:

The ratio [A-]/[HA] increase when the pH increase and the ratio decrease when the pH decrease.

Explanation:

Every weak acid or base is at equilibrium with its conjugate base or acid respectively when it is dissolved in water.

$HA + H_{2}O$ ⇄ $A^{-} + H_{3}O^{+}$

This equilibrium depends on the molecule and it acidic constant (Ka). The Henderson-Hasselbalch equation,

$pH = pKa + Log \frac{[A^{-}]}{[HA]}$

shows the dependency between the pH of the solution, the pKa and the concentration of the species. If the pH decreases the concentration of protons will increase and the ratio between A- and AH will decrease. Instead, if the pH increases the concentration of protons will decreases and the ratio between A- and AH will increase.

5 0

3 years ago

According to the following balanced equation, 2 formula units of Iron (III) Oxide (Fe2O3) can be formed by reacting 4 atoms of i

-Dominant- [34]

You have to use Avogadro's number (6.02x10^23 molecules/mole) to find the number of moles each reactant starts off with.
moles of Fe and O₂:
12 atoms/(6.02x10^23 atoms/mole)=1.99x10^-23 mol Fe
6 molecules/(6.02x10^23 molecules/mole)=9.967x10^-24 mol O₂
Then you find the limiting reagent by finding how much product each given amount of reactant can make. Which ever one produces the least amount of product is the limiting reagent.
amount of Fe₂O₃ produced:
(1.99x10^-23 mol Fe)x(2mol/4mol)= 9.967x10^-24mol Fe₂O₃
(9.967x10^-24 mol O₂)x(2mol/3mol)= 6.645x10^-24 mol Fe₂O₃
since oxygen produces the leas amount of product, oxygen is the limiting reagent. since we know that oxygen is the limiting reagent we can use the amount of product formed with oxygen to find the amount of iron used.
6.645x10^-24 mol Fe₂O₃x(4mol/2mol)=1.329x10^-23 mol Fe consumed
 find the amount left over by subtracting the original amount of Fe by the amount consumed in the reaction.
1.993x10^-23-1.329x10^-23= 6.645x10^-23mol Fe left
find the number of atoms by multiplying that by Avogadro's number.
(6.645x10^-23mol)x(6.02x10^23 atoms/mol)=4 atoms
therefore 4 atoms of Fe will be left over after the reaction happens.

I hope this helps.

8 0

2 years ago

Read 2 more answers

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0

2 years ago

A compound contains nitrogen and a metal. This compound goes through a combustion reaction such that compound X is produced from

tia_tia [17]

Answer:

The correct answer is: X is nitrogen dioxide, and Y is a metal oxide

Explanation:

Combustion of compound of containing nitrogen and metal will give nitrogen dioxide and metal oxide as product. During combustion reaction a compound reacts with oxygen in order to yield oxides of elements present in the compound.

The general equation is given as:

$4M_3N_x+7xO_2\rightarrow 4xNO_2+6M_2O_x$

Hence, the correct answer is :X is nitrogen dioxide, and Y is a metal oxide.

6 0

3 years ago