Differentiate between sandy soil, clayey soil and loamy soil on the basis of the growth of plants????

1 answer:

8 0

The mixture of rock particle sand humus is called the soil.
If soil contains greater proportion of big particles it is called sandy soil. If the proportion of fine particles is relatively higher, then it is called clayey soil. If the amount of large and fine particles is about the same, then the soil is called loamy.

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Draw the structure of 2 methyl,but-3-ene

Maslowich

Answer:

Please find the structure attached as an image

Explanation:

Based on the characteristics ending name (-ene) of the organic compound above, it belongs to the ALKENE GROUP. Alkenes are characterized by the possession of a carbon to carbon double bond (C=C) in their structure.

- But-3-ene tells us that the organic compound has four straight carbon atoms with the C=C (double bond) located on the THIRD carbon depending on if we count from right to left or vice versa.

- 2 methyl indicates that the methyl group (-CH3) is located as an attachment on the second carbon (carbon 2).

N.B: In the structure attached below, the counting is from the left to right (→).

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Because the ice helps cool it up

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The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$