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Olin [163]
3 years ago
5

PLEASE HELP!!!!!!!!!!!!

Chemistry
1 answer:
Nikitich [7]3 years ago
5 0

Answer:

1 : NA = 1    |      2 : N = 2  |     3 :  

N = 1          |       O = 5     |

O = 3         |                     |

Explanation:

im sorry but ion know the last one

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spin [16.1K]

Answer:

6 neutrons. The mass number is the number of protons and neutrons in the nucleus of an atom. The atomic number shows how many protons. Subract the atomic number from the mass number to find the number of neutrons.

Explanation:

6 0
3 years ago
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Iupac name of h3C-C-Ch2-ch3​
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Answer:

The IUPAC name of given compound is 3−5−ethyl−5−−3−methylheptane. Explanation: The parental chain is of 7 carbons with single bonds hence it is heptane. Two substituents ethyl and methyl group are attached from an equal distance. Hence according to the alphabetical order preference, counting starts from carbon which is close to an ethyl group.

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2 years ago
Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. T
Ostrovityanka [42]

Explanation:

The given data is as follows.

          \Delta H = 286 kJ = 286 kJ \times \frac{1000 J}{1 kJ}

                            = 286000 J

 S_{H_{2}O} = 70 J/^{o}K,      S_{H_{2}} = 131 J/^{o}K

 S_{O_{2}} = 205 J/^{o}K

Hence, formula to calculate entropy change of the reaction is as follows.

          \Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)

                     = [(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]

                    = [(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]

                    = 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

             \Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}

                            = 286000 J - (163.5 J/K \times 298 K)

                            = 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

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