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NikAS [45]
3 years ago
11

Draw the two-carbon hydrocarbons propane, propene, and propyne. Classify each molecule, and explain the differences between them

.
Chemistry
1 answer:
Agata [3.3K]3 years ago
5 0
Two carbon compounds have the prefix eth-, the prefix pro- applies to three carbon atoms.
Propane: CH₃CH₂CH₃, alkane molecule, completely saturated and contains no double bonds
Propene: CH₃CH=CH₂, alkene molecule, single degree of unsaturation and contains one double bond
Propyne: CH₃C≡CH, alkyne molecule, two degrees of unsaturation and contains one triple bond
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Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9.30 10-8 at 700°C. 2 H2S(g) 2 H2(g) + S2(g) If
Vikentia [17]

Answer:

The equilibrium concentration of hydrogen gas is 0.0010 M.

Explanation:

The equilibrium constant of the reaction = K_c=9.30\times 10^{-8}

Moles of hydrogen sulfide = 0.31 mol

Volume of the container = 4.1 L

[concentration]=\frac{moles}{volume (L)}

[H_2S]=\frac{0.31 mol}{4.1 L}=0.076 M

2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)

Initially

0.076 M

At equilibrium

(0.076-2x)                         2x     x

The expression of an equilibrium constant :

K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}

9.30\times 10^{-8}=\frac{(2x)^2\times x}{(0.076-x)^2}

Solving for x:

x = 0.00051

The equilibrium concentration of hydrogen gas:

[H_2]=2x=2\times 0.00051 M=0.0010 M

4 0
4 years ago
the internal energy of an ideal gas depends only on its temperature. Do a first-law analysis of this process. A sample of an ide
Aleks04 [339]

Answer:

Check Explanation

Explanation:

The Mathematical expression of the first law is given as

ΔU = Q - W

Note that depending on convention, the mathematical expression for the first law can be written as

ΔU = Q + W

a) Yes, the gas does work on its surroundings. Since it expands; indicating that there is a change in volume.

The work done is given as -PΔV if we are using the convention of (ΔU = Q + W). And this work is negative work done, since the system expands and does work on the surroundings.

But if we are using the convention of (ΔU = Q - W), the work done is given as PΔV and the workdone by the system on the environment is positive work during expansion.

b) Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.

So, from the mathematical expression,

ΔU = Q + W or ΔU = Q - W

If ΔU = 0,

Q = - W or Q = W

But either ways,

Q = PΔV

(because, W = -PΔV for ΔU=Q+W and W = PΔV for ΔU=Q-W)

So, either ways, the heat transfer is the same and it is positive. This indicates heat is transferred from the surroundings to the system.

(c) What is ΔU for the gas for this process?

Since this all takes place at constant temperature and against a constant atmospheric pressure, the change in internal energy for this system is 0. Change in internal energy depends on a change in temperature.

Hope this Helps!!!

8 0
4 years ago
Explain how soil can affect the composition of the solution that move through it
sertanlavr [38]

Answer: Soils are formed through the interaction of five major factors: time, climate, parent material, topography and relief, and organisms.

Soil structure affects plant growth in many, often surprising, ways. The most obvious effects are on root growth, which is strongly inhibited by hard soil, and which in turn influences the ability of the root system to extract adequate water and nutrients from the soil.

Explanation:

https://www.publish.csiro.au/sr/pdf/SR9910717

This is where I gather some info.  

5 0
3 years ago
A quantity of water vapor at 100​°​C is condensed to liquid water at the same temperature. In the process, 180,414.08 joules of
salantis [7]

Answer:

0.08kg

Explanation:

Given parameters:

Temperature of water  = 100°C

Quantity of heat released = 180414.08J

Unknown:

Quantity of water condensed  = ?

Solution:

This is simple phase change process without any attendant change in temperature.

To solve this problem, use the formula below;

             H  = mL

H is the heat released

m is mass

L is the latent heat of condensation of water  = 2260 kJ/kg

 Insert the parameters and solve;

       180,414.08 = m x  22.6 x 10⁵ J/kg

         m  = \frac{180414.08}{2260000}   = 0.08kg

5 0
3 years ago
suppose you mix 100.0 g of water at 22.6 oc with 75.0 g of water at 75.4 oc. what will be the final temperature of the mixed wat
Arada [10]

The temperature of mixed water is 45.230C under the conditions stated.

<h3>How can the change in temperature when mixing water be calculated?</h3>

T(final) = (m1 T1 + m2 T2) / (m1 + m2), at which m1 and m2 are indeed the weight training of the water in the initial and 2nd canisters, T1 is the water temperature in the first container, and T2 is the water temperature in the second container, can be used to determine the final the water's temperature mixture.

<h3>Briefing:</h3>

The given parameters;

mass of the cold water, m = 100 g

initial temperature of the water, t₁ = 22.6 ⁰C

initial temperature of the hot water, t₂ = 75.4⁰ C

75 g is the hot water's mass.

specific heat capacity of water is 4.184 J/g⁰C

The mixture's final temperature is estimated as follows;

Based on the principle of conservation of energy;

Heat received by the ice water equals heat lost by the hot water.

mcΔθ₂ = mcΔθ₁

75 x 4.184 x (75.4 - T) = 100 x 4.184 x (T - 22.6)

75 x (75.4 - T) = 100 x (T - 22.6)

(75.4 - T) = 1.333(T - 22.6)

75.4 - T = 1.333T - 30.1258

75.4 + 30.1258 = 1.333T + T

105.5258 = 2.333T

T = 45.2318⁰C

To know more about temperature visit :

brainly.com/question/29386637

#SPJ4

5 0
2 years ago
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