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3 years ago

How many neutrons and electrons are in Oxygen⁻¹⁷ isotope?

Chemistry

1 answer:

Ivenika [448]3 years ago

7 0

Answer:

Neutrons- 9 Electrons- 8

Explanation:

Protons equals electrons and since the atomic# of oxygen is 8, 8 is the proton and 8 will be the electrons a s well since it isn't an ion. You subtract 8 from the 17 to get 9 as the neutron.

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Consider the following reaction at constant P. Use the information here to determine the value of Î”Ssurr at 355 K. Predict whet

Sloan [31]

The given reaction is as follows:

2NO (g) + O₂ (g) = 2NO₂ (g), ΔH = -114 kJ

It is known that dSsurr = -dHsys / T (Temp = 355 K)

So, dSsurr = - (-114 × 1000) / 355

dSsurr = +321.12 J/K

Hence, the value of dSsurr is +321.12 J/K

For a reaction to be spontaneous, dG<0,

Also dStotal = dSsys + dSsurr > 0

It is known that dG = dHsys - TdSsys,

Now let us assume,

dG<0

Also, dStotal = dSsys + dSsurr > 0

(-114 × 1000) - (355 × dSsys) <0

355 × dSsys > -114 × 1000

dSsys > -321

dSsys >dSsurr

dSsys + dSsurr > 0

dStotal > 0

Thus, the assumption is correct, and the given reaction is spontaneous. Hence, the final answer is Ssurr = +321 J/K reaction is spontaneous.

8 0

3 years ago

What is the difference between primary scale and extension scale?

elena-14-01-66 [18.8K]

Answer:

8txirzrz8rzrirzriz£=,£=,,-8/-9/-¥/

Explanation:

otcitxfi

8 0

3 years ago

Give the name and formula of the compound formed from the following elements: a) Sodium and nitrogen b) Oxygen and strontium c)

elena-s [515]

Answer: a) Sodium and nitrogen : $NaN_3$ : sodium nitride.

b) Oxygen and strontium : $SrO$ : strontium oxide.

c) Aluminum and chlorine : $AlCl_3$ : aluminium chloride.

d) Cesium and bromine : $CsBr$ : cesium bromide.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

a) Sodium and nitrogen : Here sodium is having an oxidation state of +1 called as $Na^{+}$ cation and nitrogen forms an anion $N^{3-}$ with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $NaN_3$ named as sodium nitride.

b) Oxygen and strontium : Here strontium is having an oxidation state of +2 called as $Sr^{+}$ cation and oxygen forms an anion $O^{2-}$ with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $SrO$ named as strontium oxide.

c) Aluminum and chlorine : Here aluminium is having an oxidation state of +3 called as $Al^{3+}$ cation and chlorine forms an anion $Cl^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $AlCl_3$ named as aluminium chloride.

d) Cesium and bromine : Here cesium is having an oxidation state of +1 called as $Cs^{+}$ cation and bromine forms an anion $Br^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $CsBr$ named as cesium bromide.