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3 years ago

There are σ bonds and π bonds in h3c-ch2-ch=ch-ch2-c≡ch.

Chemistry

1 answer:

FrozenT [24]3 years ago

6 0

The number of sigma and pi bonds are,

Sigma Bonds = 16

Pi Bonds = 3

Explanation:
Every first bond formed between two atoms is sigma. Pi bond is formed when already a sigma bond is there. While in case of Alkyne (triple Bond) there is one sigma and one pi bond already present, so the third bond is formed by second side-to-side overlap of orbitals, hence, a second pi bond is formed.
Below all black bonds are sigma bonds, while in alkene there is one pi bond and in alkyne there are two pi bonds.

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When bridges are built, special joints must be used because the material of the bridge shrinks, and without these joints, the ma

marysya [2.9K]

Expansion

Explanation:

Expansion is the change is size of a body with increasing temperature and tension.

Bridges are built of iron. Iron thermally expands in the presence of increased temperature during the day. The length of the iron will increase by simple fractions. At night, when temperature drops, the bar returns back to its initial position.

The allowance for expansion of the metal is made in order to prevent tension within the structure.
Not doing this will cause the material to break since there is no free space for the bar to change shape.

Learn more:

Expansion in gases brainly.com/question/9979757

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6 0

3 years ago

Find the density of a cube on Earth that weighs 1.5 kg and has a side-length of 10 cm.

tamaranim1 [39]

Answer:

1500kg/m^3

Explanation:

Formula:

d=m/v

Given:

m=1.5kg

v=1000cm^3

(The side length of a cube is always equal to the others)

Required:

d=?

Solution:

d=m/v

d=1.5kg/1000cm^3

d=1.5kg/0.001m^3

d=1500kg/m^3

Hope this helps ;) ❤❤❤

3 0

3 years ago

If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc

hodyreva [135]

Answer: 69.72 kg of cryolite will be produced.

Explanation:

The balanced chemical equation is:

$Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $Al_2O_3$ = $\frac{16.9\times 1000g}{102g/mol}=165.7moles$

moles of $NaOH$ = $\frac{57.4\times 1000g}{40g/mol}=1435moles$

moles of $HF$ = $\frac{57.4\times 1000g}{20g/mol}=2870moles$

As 1 mole of $Al_2O_3$ reacts with 6 moles of $NaOH$

166 moles of $Al_2O_3$ reacts with = $\frac{6}{1}\times 166=996$ moles of $NaOH$

As 1 mole of $Al_2O_3$ reacts with 12 moles of $HF$

166 moles of $Al_2O_3$ reacts with = $\frac{12}{1}\times 166=1992$ moles of $HF$

Thus $Al_2O_3$ is the limiting reagent.

As 1 mole of $Al_2O_3$ produces = 2 moles of cryolite

166 moles of $Al_2O_3$ reacts with = $\frac{2}{1}\times 166=332$ moles of cryolite

Mass of cryolite $(Na_3AlF_6)$ = $moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg$

Thus 69.72 kg of cryolite will be produced.

8 0

3 years ago

Given K = 3.61 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.19 at 45°C for the reaction 2 A(g)

Firlakuza [10]

Answer:

K = 0.55

Kp = 0.55

mol fraction B = 0.27

Explanation:

We need to calculate the equilibrium constant for the reaction:

C(g) + D(g) ⇄ 2B(g) K₁= ? (1)

and we are given the following equilibria with their respective Ks

A(g) + B(g) ⇄ C(g) K₂= 3.61 (2)

2 A(g) + D(g) ⇄ C(g) K₃= 7.19 (3)

all at 45 ºC.

What we need to do to solve this question is to manipulate equations (2) and (3) algebraically to get our desired equilibrium (1).

We are allowed to reverse reactions, in that case we take the reciprocal of K as our new K' ; we can also add two equilibria together, and the new equilibrium constant will be the product of their respective Ks .

Finally if we multiply by a number then we raise the old constant to that factor to get the new equilibrium constant.

With all this in mind, lets try to solve our question.

Notice A is not in our goal equilibrium (3) and we want D as a reactant . That suggests we should reverse the first equilibria and multiply it by two since we have 2 moles of B as product in our equilibrium (1) . Finally we would add (2) and (3) to get (1) which is our final goal.

2C(g) ⇄ 2A(g) + 2B(g) K₂´= ( 1/ 3.61 )²

₊

2 A(g) + D(g) ⇄ C(g) K₃ = 7.19

C(g) + D(g) ⇄ 2B(g) K₁ = ( 1/ 3.61 )² x 7.19

K₁ = 0.55

Kp is the same as K = 0.55 since the equilibrium constant expression only involves gases.

To compute the last part lets setup the following mnemonic ICE table to determine the quantities at equilibrium:

pressure (atm) C D B

initial 1.64 1.64 0

change -x -x +2x

equilibrium 1.64-x 1.64- 2x

Thus since

Kp =0.55 = pB²/ (pC x pD) = (2x)²/ (1.64 -x)² where p= partial pressure

Taking square root to both sides of the equation we have

√0.55 = 2x/(1.64 - x)

solving for x we obtain a value of 0.44 atm.

Thus at equilibrium we have:

(1.64 - 0.44) atm = 1.20 atm = pC = p D

2(0.44) = 0.88 = pB

mole fraction of B = partial pressure of B divided into the total gas pressure:

X(B) = 0.88 / ( 1.20 + 1.20 + 0.88 ) = 0.27

8 0

3 years ago

Blababa [14]

Mtge answer is a malaria

5 0

3 years ago

Read 2 more answers

There are ________ σ bonds and ________ π bonds in h3c-ch2-ch=ch-ch2-c≡ch.

There are σ bonds and π bonds in h3c-ch2-ch=ch-ch2-c≡ch.