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4 years ago

Is air a compound mixture or a element.

Chemistry

2 answers:

GenaCL600 [577]4 years ago

8 0

Is air a compound mixture or a element. compound

adelina 88 [10]4 years ago

6 0

Air is not a mixture because of scientists freezing it and finding different liquids, it is a mixture because the compounds that make up air e.g. oxygen (o2), Carbon dioxide (co2) and the most important Nitrogen which is an element and makes up 78.09% of air are not chemically bound in the way that compounds are ...

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The solubility of acetanilide is 12.8 g in 100 mL of ethanol at 0 ∘C, and 46.4 g in 100 mL of ethanol at 60 ∘C. What is the maxi

weqwewe [10]

Answer: 72.41% and 26.90% respectively.

Explanation:

At 60°C, you can dissolve 46.4g of acetanilide in 100mL of ethanol. If you lower the temperature, at 0°C, you can dissolve just 12.8g, which means (46.4g-12.8g)=33.6g of acetanilide must have precipitated from the solution.

We can calculate recovery as:

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{33.6\ g}{46.4\ g}*100=72.41\%$

So the answer to the first question is 72.41%.

For the second part just use the same formula, the mass of the precipitate is the final mass minus the initial mass, (171mg-125mg)=46mg.

$\%R=\frac{crystalized\ mass}{initial\ mass}*100 =\frac{46\ mg}{171\ mg}*100=26.90\%$

So the answer to the second question is 26.90%.

3 0

3 years ago

A water sample from a hot thermal vent contained a single-celled organism that had a cell wall but lacked a nucleus. What is its

a_sh-v [17]

Answer:

Archaea

Explanation:

Archaea are prokaryotic single cell microorganism.

They are usually found in extreme environment like hot thermal vent and others. They share characteristics with bacteria and eukaryote. Archaea as a prokaryotic organism they lack true nucleus and organelles but unlike bacteria they are unaffected by antibiotics and contain different cell wall components. Unlike bacteria and eukaryotes, their membranes contain branching lipids.

6 0

3 years ago

How many formula units are contained in 2.35mol of Iron (III) Oxide

konstantin123 [22]

Remember, 1 mole= 6.022x10^23 atoms, molecules, or formula units.

Answer is 1.42x10^24

7 0

3 years ago

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e

sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer: The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

Explanation:

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$

Initial: 1.8 4.7

At eqllm: 1.8-x 4.7-x

Evaluating the value of 'x'

$\Rightarrow (1.8-x)=1.16\\\\x=0.64$

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}$

$p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm$

Putting values in above expression, we get:

$K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136$

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$

Initial: 2.36 4.06 0.64

At eqllm: 2.36-x 4.06-x 0.64+x

Putting value in the equilibrium constant expression, we get:

$0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41$

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

3 0

3 years ago

The concentrations of the products at equilibrium are [pcl3] = 0.180 m and [cl2] = 0.250 m. what is the concentration of the rea

sineoko [7]

We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula:
Kc = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 = 0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07

4 0

3 years ago