Question

Once you have filled in all the gaps on a separate sheet of paper, enter the four missing element symbols in the first row of the table into the answer box below.

Learning Goal: This question reviews the format for writing an element's written symbol. Recall that written symbols have a particular format. Written symbols use a form like this: CI In this form the mass number, 35, is a stacked superscript. The atomic number, 17, is a stacked subscript. "CI" is the chemical symbol for the element chlorine. A general way to show this form is: mass number Symbol atomic number It is also correct to write symbols by leaving off the atomic number, as in the following form: 35 Cl or mass number Symbol This is because if you write the element symbol, such as Cl, you know the atomic number is 17 from that symbol. Remember that the atomic number, or number of protons in the nucleus, is what defines the element. Thus, if 17 protons are in the nucleus, the element can only be chlorine. Sometimes you will only see 35 Cl, where the atomic number is not written. Watch this video to review the format for written symbols. In the following table each column represents a neutral atom. Fill in the gaps in the following table by working this problem out on a separate sheet of paper. You will need the periodic table in the "resources" drop down menu in the upper right hand corner of this question to look up atomic numbers and masses. 79 Br Symbol Protons 26 50 Neutrons 30 64 Electrons 48 54 Mass no. 132 119 Use the hints in the following part whenever needed to review the steps to filling in this type of table.

Answer 1

Page No. Date -0. Firstly_giren BV Therefore mass number of Bromine is 79 since each column represents a neutral atom Therefore no. of electrons = mo- of protons And we know No. of protons = Atomic numbex of atom... for Bromine, Atomic number = 35 Therefore number of poroton = number of electrons a = patent Atomic number = 35 to find number of neutrons : ANOS elips.si in un Mass number - No. of Protons = 79-35 ICBecause mass number = 30. of protons t no. of neutrons) 34 fu column I No. of protons = 26 No. of neutrons = 30 1. Mass no. = No. of parton's+ 30. of neutrons 26 +30 23 56 No. of electrons = 26. 5. Atom having atomic no 26 is loon cred

Page No. Date (1 column (WI) I No. of electrons = 48 No.of protons 248_ I No. of neutrons = 64 1. Mlass number = 64+48 112 o Alom having atomic number 48 is Cadmium (Cd) 1 Column (A) in No. of electrons = now of protons = 54 Mass no. = 132. mass 70.-70.0 portons 11:. 1499€ No. of neutrons = me of protons maileno _ Birgit ir = 132-54 on => 78 Atom having 54 atenuc no. is_Xenon cxe) v Column (A) No. of protons = no. of electrons = 50 - Mass no. =119 No. of neutrons = mass mo. - no. of poplans 5) 119-50 2 69 Atom having atomic no. 5o is Tim (sn)

Date Il Protons | 35 | 26 148 154 150 I Neutrons| 34 | 30 | 64 | 78 169 1 1 Electrons 1 35 126 1 48 154 150 Mass no. 1 79 156 | 112 | 132 | 119 This column wise : - - Tysn THB Soc. 112 132 119 Brzaste, ugld a guxe 150