Which set of changes does not involve breaking or forming chemilcal bonds

Chemistry

1 answer:

sineoko [7]3 years ago

6 0

<h2><u>Answer:</u></h2>

It wasn't an adjustment in the condition of issue on the grounds that the vitality in the can did not change. Additionally, since this was a physical change, the atoms in the can are as yet similar particles. No synthetic bonds were made or broken. You added enough vitality to make a stage change from strong to fluid.

The main changes recorded which don't include framing or breaking substance bonds would bubble and liquefying. Bubbling and liquefying are physical changes as opposed to synthetic changes, so no new items are shaped.

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6 0

2 years ago

Given the balanced equation representing a reaction:

Vladimir79 [104]

Answer:

Option A. Addition

Explanation:

Unsaturated compounds under goes addition reaction to produce saturated compounds..

In the equation given above i.e

H2C=CH2 + F–F —> FCH2CH2F

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3 years ago

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 $H^+$