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4 years ago

How many mono-; di- ; and trichloro derivatives are possible for cyclopentane?

Chemistry

1 answer:

Studentka2010 [4]4 years ago

8 0

Mono - 1 Di- 2 Tri - 4

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What is silicons reactivity level

egoroff_w [7]

Answer:

this is your answer

Explanation:

Silicon, like carbon, is relatively inactive at ordinary temperatures; but when heated it reacts vigorously with the halogens (fluorine, chlorine, bromine, and iodine) to form halides and with certain metals to form silicides.

hope it helps

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5 0

3 years ago

Read 2 more answers

Why are halogens reactive?

erica [24]

<u>Halogens are reactive because:</u>

They have high electronegativity and also high nuclear charge. So, they are reactive and also gain an electron when they react with other elements.

Since they are very reactive, halogens are very harmful to living organisms. Some of the halogens are fluorine, chlorine, bromine, iodine, astatine. These are mostly non metals. Fluorine is one of the most reactive gas and also very toxic gas. When Fluorine reacts with glass along with small amounts of water, it forms silicon tetra fluoride (SiF4). Hence fluorine should be handled with substances like the inert organofluorine compound Teflon.

8 0

3 years ago

Which is true of ultraviolet rays?

Ket [755]

A is true of UV rays.
B is true not of UV rays but rather of visible light.
C is true not of UV rays but rather of microwaves. (unless you actually toast your toast in a toaster like a normal person)
D is true not of UV rays but rather of radio waves.

7 0

3 years ago

One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc

Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

$\text{The new partial pressure for }N_2 \ gas}$

$P_1V_1=P_2V_2$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar$

$\text{The new partial pressure for }Ar \ gas}$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar$

$Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.225 \ Bar$

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

$2.1 \ bar = 2.07 \ atm \\ \\3.4 \ bar = 3.36 \ atm$

For moles N₂:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}$

$n = 0.08297 \ mol \ N_2$

For moles of Ar:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}$

$n = 0.2172 \ mol \ Ar$

$\mathtt{total \ moles = moles \ of \ N_2 + moles \ of \ Ar}$

$=0.08297 mol + 0.2037 mol \\ = 0.2867 mol gases$

Finally;

The final pressure of the mixture is:

$PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}$

P = 2.217 atm

P ≅ 2.24 bar

7 0

3 years ago

In Ontario, some electricity comes from coal-burning generators. Coal is a natural form of carbon that has a large amount of sul

irinina [24]

Answer:

$C(s)+O_{2}(g) \rightarrow CO_{2}(g)$

$S(s)+O_{2}(g) \rightarrow SO_{2}(g)$

Sulphurdioxide

$SO_{2}(g) + H_{2}O(l) \rightarrow H_{2}SO_{3}(l)$

$SO_{2}(g) + \frac{1}{2}O_{2} \rightarrow SO_{3}(g)$

Incomplete combustion produce harmful gases.

Explanation:

Carbon reacts with atmospheric oxygen to form carbondioxide as well as sulphur dioxide.

The chemical equations are as follows.

$C(s)+O_{2}(g) \rightarrow CO_{2}(g)$

$S(s)+O_{2}(g) \rightarrow SO_{2}(g)$

Sulfur dioxide is very harmful to the environment to cause acid rains.

This harmful gas mix with rain water to form sulphuric acid.

The balanced chemical equation for the reaction that produces harmful environmental effects is as follows.

$SO_{2}(g) + H_{2}O(l) \rightarrow H_{2}SO_{3}(l)$

$SO_{2}(g) + \frac{1}{2}O_{2} \rightarrow SO_{3}(g)$

$SO_{3}(g) +H_{2}O(l) \rightarrow H_{2}SO_{4}(l)$

In the absence of proper amount of oxygen required for combustion, incomplete combustion will take place which will result in formation of more carbondioxide an other harmful gases.

4 0

3 years ago