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Nesterboy [21]
3 years ago
7

3. Make an argument for placing hydrogen in the halogen group rather than the alkali metals. 4. Compare and contrast ionization

energy and atomic radius. 5. When the element with the atomic number of 119 is discovered, what group will it be in? Explain your logic. 6. A new element is discovered that is very unstable and highly reactive, and it likes to lose its one valence electron. In what group should this element be placed in? Explain.
Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
6 0

Explanation:

Question 3

Hydrogen is an element with a single electron and proton. It has no neutron.

  • The electronic configuration of hydrogen is given as 1s¹
  • The maximum of number of electrons in the S-subshell is just 2.
  • In order to complete its octet and be like the closest noble gas which is helium, it must be willing to accept one more electron.
  • The halogens in group 7 also have this unique property. They all have 7 electrons in their valence shell. This implies that they only need just an electron to complete their octet.
  • This shared property of hydrogen and the halogens makes for a strong reason for the element to be in group 7.

Learn more:

halogens brainly.com/question/1380547

#learnwithBrainly

--------------------------------------------------------------------------------------------------------------

Question 4

Comparison:

  • Ionization energy is a measure of the readiness of an atom to lose an electron. The lower the value, the easier it is for an atom to lose an electron and vice versa.
  • The atomic radius is the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state of metals.
  • The atomic radius like the ionization are periodic trends and they tell about the nature of elements on the periodic table.

Contrast:

  • Atomic radius is estimated using two nuclei of an atom but ionization is determined for a single element.
  • Atomic radius increases down a group on the periodic table whereas ionization energy decreases down a group.
  • Across a periodic table, atomic radii decreases progressively from left to right but ionization energy increases across a period.

Learn more:

calculations using atomic radius brainly.com/question/5048216

Ionization energy brainly.com/question/6324347

#learnwithBrainly

--------------------------------------------------------------------------------------------------------------

Question 5

The element when discovered will be in group 1. This is the group of the alkali metals. These metals are highly reactive.

  • The shell structure of the element will be 2,8 18,32,32,18,8,1
  • A neutral atom will have a notation of  [Og] 8s¹
  • This further implies that it will be in the s-block.
  • The outer electrons are useful in determining the periodic groups of elements.
  • For elements in group 1, they all have 1 valence electron. Those in group 8 have eight outermost electron.

Learn more:

periodic table brainly.com/question/2014634

#learnwithBrainly

--------------------------------------------------------------------------------------------------------------

Question 6

The element will be group 1 as an alkali metal.

  • Group 1 elements are known for their high reactivity and their willingness to loose their valence electron.
  • Elements gain, lose or share electrons in order to attain stability.
  • The loss of the element's valence electron confers a special stability on them and makes them achieve an octet configuration.
  • Therefore the element will fit perfectly well in group 1.

Learn more:

periodic functions  brainly.com/question/11730855

#learnwithBrainly

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A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The
natima [27]
1) Compund Ir (x) O(y)

2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

5) Convert grams to moles

moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

6) Find the proportion of moles

Divide by the least of the number of moles, i.e. 0.00656

Ir: 0.00656 / 0.00656 = 1

O: 0.0131 / 0.00656 = 2

=> Empirical formula = Ir O2 (where 2 is the superscript for O)

Answer: Ir O2
5 0
3 years ago
Read 2 more answers
The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1
Harman [31]

Answer:

54 days

Explanation:

We have to use the formula;

0.693/t1/2 =2.303/t log Ao/A

Where;

t1/2= half-life of phosphorus-32= 14.3 days

t= time taken for the activity to fall to 7.34% of its original value

Ao=initial activity of phosphorus-32

A= activity of phosphorus-32 after a time t

Note that;

A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)

Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

0.048=2.6/t

t= 2.6/0.048

t= 54 days

3 0
3 years ago
Predict the products of La(s) + O2(aq) ->
mina [271]

Answer:

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

Explanation:

<em>Hello </em><em>there!</em>

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

From the above question I have considered that my La(s), solid is either Sodium (Na) - group I, Magnesium - group II and Aluminum - group III.

In a reaction, there is exchange of electrons given by their oxidation numbers (I, II and III - for our metals above)

The chemical equations are thus;

i. Na (s) + O2 (g) => NaO (s)

ii. Mg (s) + O2 (g) => Mg2O (s)

iii. Al (s) + O2 (g) => Al2O3 (s)

Relate this to the problem and it will be;

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

<em>I hope this </em><em>helps </em><em>you</em><em> </em><em>to </em><em>understand</em><em> </em><em>better</em><em>.</em><em> </em><em>Enjoy </em><em>your</em><em> </em><em>studies</em>

3 0
2 years ago
What is the molar mass of a substance
aleksandrvk [35]
<span>47.88 g/mol is the awsner your welcome</span>
3 0
3 years ago
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Be sure to answer all parts.The combustion of a 40.0−g gaseous mixture of H2 and CH4 releases 3766 kJ of heat. Calculate the amo
Svetach [21]

Answer:H2=11.4g

CH4=28.6g

Explanation:The complete combustion of the two gases can be represented by a balanced reaction below

1. CH4 +2O2___CO2+2H2O

2.2H2+O2___2H2O

Combining the two we have CH4 +2H2+3O2___

CO2+4H2O

Since the mixture contains 40gof CH4 and 2, therefore 20g of CH4 and 8g of H2 combines.

Calculated from their molecular Mass i.e CH4 12+4×2)=20 and 2H2= 2×2×2=8g

Mass of CH4=20/28×40=28.6g

2H2=8/28×40=11.4g

7 0
3 years ago
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