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Finger [1]
3 years ago
10

A substance that has high reactivity

Chemistry
2 answers:
Fiesta28 [93]3 years ago
5 0
A. easily combines chemically with other substances is your answer.
vichka [17]3 years ago
5 0
A substance that has high reactivity <span>A. easily combines chemically with other substances.</span><span>
</span>
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2 SO3 (g) + Heat &lt;-----&gt; 2 SO2 (g) + O2 (g)
Step2247 [10]

Answer:

The concentration of SO₂ will decreases

Explanation:

As you can see in the reaction

2 moles of gas ⇆ 3 moles of gas

Based on Le Châtelier's principle, a change doing in a system will produce that the system reacts in order to counteract the change made.

If the pressure is increased, the system will shift to the left in order to produce less moles of gas and decrease, thus, the pressure.

As the system shift to the left, the concentration of SO₂ will decreases

7 0
3 years ago
What is Catenation in chemistry?​
Eddi Din [679]

Answer:

In chemistry, catenation is the bonding of atoms of the same element into a series, called a chain. A chain or a ring shape may be open if its ends are not bonded to each other, or closed if they are bonded in a ring

Explanation:

5 0
3 years ago
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Is a television remote ac or dc
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Dc like a battery mate
3 0
3 years ago
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
3 years ago
How many liters are in 3.45 moles of oxygen gas at STP?
just olya [345]

Answer:

22.4L of one mole of any gas

or you can use PV=nRT

3.45*22.4=77.28

Explanation:

5 0
2 years ago
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