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sasho [114]
3 years ago
6

the isotope of an atom containing 31 protons and 39 neutrons suddenly has two neutrons added to it. what isotope is created? a.

yttrium-41 b. yttrium-72 c. gallium-41 d. gallium-72
Chemistry
2 answers:
olasank [31]3 years ago
7 0
The elements are identified by the number of protons of the atom, which is its atomic number.

In this case the number of protons 39 (atomic number 39) permit you to identify the element as gallium.

Now, to identify the isotope you tell the name of the element and add the mass number.

The mass number is the sum of the protons and the neutrons

In this case, the number of neutrons is the original 39 plus the 2 added suddenly, i.e. 39 + 2 = 41, so the mass number is 31 + 41 = 72

Therefore, the isotope is gallium - 72.

Answer: isotope gallium - 72

trasher [3.6K]3 years ago
4 0
The answer is D. Gallium-72.


Hope this helps:)
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Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat
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Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values ​​of the atomic mass of each element that form the compounds:

  • N: 14 g/mol
  • H: 1 g/mol
  • O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

  • NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
  • O₂: 2*16 g/mol= 32 g/mol
  • NO: 14 g/mol + 16 g/mol= 30 g/mol
  • H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

  • NH₃: 4 moles
  • O₂: 5 moles
  • NO: 4 moles
  • H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

  • NH₃: 4 moles*17 g/mol= 68 g
  • O₂: 5 moles*32 g/mol= 160 g
  • NO: 4 moles*30 g/mol= 120 g
  • H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }

mass of O₂≅17.65 g

<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>

3 0
3 years ago
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