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1 year ago

How are alkanes, alkenes and alkynes different?

Chemistry

1 answer:

Alenkasestr [34]1 year ago

5 0

The difference between alkanes, alkenes and alkynes comes from its structure.

Al of them are composed only by carbon and hydrogen atoms, but they differ in the type of bond between carbon atoms.

An alkane have only single bonds between carbon atoms, and for this they are called saturated. Here is an example:

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During protein synthesis, the covalent bond between amino acids is formed by the process of _____.

kati45 [8]

Dehydration synthesis

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4 years ago

Be sure to answer all parts. What can be added to the equilibrium AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl−(aq) (a) that will shi

lora16 [44]

Answer:

a) Ag(NH₃)₂⁺, Cl⁻.

b) NH₃.

c) AgCl.

Explanation:

Based on LeChatelier's law, a system in chemistry can change responding to a disturbance of concentration, temperature, etc. in order to restore a new state.

In the reaction:

AgCl(s) + 2NH₃(aq) ⇌ Ag(NH₃)₂⁺(aq) + Cl⁻(aq)

When reactants are added, the system will produce more products restoring the equilibrium and vice versa. A reactant in solid state doesn't take part in the equilibrium, thus:

a) Ag(NH₃)₂⁺, Cl⁻. The addition of products will shift the equilibrium to the left

b) NH₃. The addition of reactant will shift the equilibrium to the right.

c) As AgCl is in solid phase, will not shift the equilibrium in either direction.

6 0

4 years ago

The rate of formation of C in the reaction 2 A + B →2 C + 3 D is 2.7 mol dm−3 s −1 . State the reaction rate, and the rates of f

Zielflug [23.3K]

Answer:

Rate of reaction = $1.35mol.dm^{-3}.s^{-1}$

Rate of consumption of A = $2.7mol.dm^{-3}.s^{-1}$

Rate of consumption of B = $1.35mol.dm^{-3}.s^{-1}$

Rate of formation of D = $4.15mol.dm^{-3}.s^{-1}$

Explanation:

According to laws of mass action for the given reaction,

$Rate= -\frac{1}{2}\frac{\Delta [A]}{\Delta t}=-\frac{\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}=\frac{1}{3}\frac{\Delta [D]}{\Delta t}$

where, $-\frac{\Delta [A]}{\Delta t}$ is rate of consumption of A, $-\frac{\Delta [B]}{\Delta t}$ is rate of consumption of B, $\frac{\Delta [C]}{\Delta t}$ is rate of formation of C and $\frac{\Delta [D]}{\Delta t}$ is rate of formation of D

Here $\frac{\Delta [C]}{\Delta t}=2.7mol.dm^{-3}.s^{-1}$

So, Rate of reaction = $(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}$

Rate of formation of D = $(\frac{3}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{3}{2}\times 2.7mol.dm^{-3}.s^{-1})=4.15mol.dm^{-3}.s^{-1}$

Rate of consumption of A = $(\frac{2}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{2}{2}\times 2.7mol.dm^{-3}.s^{-1})=2.7mol.dm^{-3}.s^{-1}$

Rate of consumption of B = $(\frac{1}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}$

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Explanation:

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