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3 years ago

Given that oxygen has an atomic number of 8.

Chemistry

2 answers:

Tcecarenko [31]3 years ago

7 0

Answer:

Oxygen (O), for example, has three naturally occurring isotopes, which can be written as 8/16O, 8/17 O, and 8/18O.

Explanation:

professor190 [17]3 years ago

4 0

Answer: 8 protons and 10 neutrons

Explanation:

Atomic number is defined as the number of protons or number of electrons that are present in an atom.

Atomic number = Number of electrons = Number of protons

atomic number = 8 = Number of protons

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons

Mass number = 18 = Number of protons + Number of neutrons

Thus Number of neutrons =18 - Number of protons= 18 - 8 = 10

Thus there are 8 protons and 10 neutrons in Oxygen -18.

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3 years ago

Oxygen always has an oxidation number of -2 unless it is combined with fluorine or found in the compound peroxide. When in the c

Tpy6a [65]

When oxygen is found is peroxide, it has an oxidation number of -1.

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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 $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

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2 years ago

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