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Naily [24]
4 years ago
13

How will the following system react when an increase in pressure occurs? Note: all species are gases. N2O4 <--> 2NO2 Quest

ion 9 options:
a. More N2O4 is produced.
b. There is no way to predict how the increase in pressure will affect this system without more information.
c.No change occurs.
d.More NO2 is produced.
Chemistry
2 answers:
Butoxors [25]4 years ago
6 0
<span>a. More N2O4 is produced.

As all species are gases, over more pressure, the side of the reaction with fewer gas molecules will be favoured as it is the one that occupies less space, where more molecules would fit in lesser space. N</span>₂O₄ has the same number of atoms of 2NO₂, but the difference is that they are only in one molecule - occupy less space because all the atoms are closer to each other.
Sophie [7]4 years ago
3 0

a is incorrect just took the test

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Force necessary to change an object's motion
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Yes force is necessary
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3 years ago
A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon
Anna35 [415]

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

3 0
3 years ago
What unit is not a measured volume
kodGreya [7K]
The base units of length and volume are linked in the metric system. By definition, a liter<span> is equal to the volume of a cube exactly 10 </span>cm<span> tall, 10 </span>cm<span> long, and 10 </span>cm<span> wide. Because the volume of this cube is 1000 cubic </span>centimeters<span> and a </span>liter<span> contains 1000 milliliters, 1 milliliter is equivalent to 1 cubic centimeter.</span>
4 0
3 years ago
Solid calcium carbonate breaks down into carbon dioxide gas, oxygen gas, and solid calcium, write an equation for described reac
Savatey [412]

Answer:

CaCO₃ ------> CaO + CO₂

<em>Note: The breakdown products of solid calcium carbonate is not carbon dioxide gas, oxygen gas, and solid calcium. Rather, it is calcium oxide and carbon dioxide as shown in the reaction equation above.</em>

Explanation:

Calcium carbonate is a mineral which occurs in nature in rocks as calcite and limestone. It is also the main component of eggshells, snail shells, seashells and pearls. It has the molecular formula CaCO₃.

Calcium carbonate or limestone decomposes into calcium oxide and carbon dioxide when heated and the reaction is used to make quicklime and carbon dioxide gas. The equation of the reaction is given below:

CaCO₃ ------> CaO + CO₂

Calcium oxide, one of the products of The decomposition reaction is known as lime and is an important mineral used for many purposes such as reducing the acidity of soils, in the production of limelight, as well as serving as the main ingredient in cement.

The main use of calcium carbonate is in the construction industry as an ingredient of cement. Calcium carbonate is also an important mineral required by sea organisms for making their shells.

5 0
3 years ago
A 3.0-L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane
zhenek [66]

Explanation:

It is known that rate of effusion of gases are inversely proportional to the square root of their molar masses.

And, half of the helium (1.5 L) effused in 24 hour. So, the rate of effusion of He gas is calculated as follows.

          \frac{1.5 L}{24 hr}

            = 0.0625 L/hr

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Now,

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or, rate of O_{2} = \frac{\text{Rate of He}}{2.83}

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This means that 0.022 L of O_{2} gas effuses in 1 hr

So, time taken for the effusion of 1.5 L of O_{2} gas is calculated as follows.

         \frac{1.5 L}{0.022 L/hr}

                = 68.18 hour

Thus, we can conclude that 68.18 hours will it take for half of the oxygen to effuse through the membrane.

3 0
4 years ago
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