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3 years ago

11

Where on the periodic table are you most likely to find elements that do not react with other elements A) group 1 B) group 2 C)

group 17 D) group 18

1 answer:

Troyanec [42]3 years ago

6 0

Answer:

d) group 18

Explanation:

Group 18 on the periodic table consists of noble gasses, which do not chemically interact with other elements.

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If the half-life of hydrogen-3 is 11.8 years, after two half-lives the radioactivity of a sample will be reduced to one-half of

maw [93]

Answer:

False

Explanation:

Half life is the time period at which the concentration of the radioactive substance in decay reduced to half.

<u>Thus, if the hydrogen-3 has gone 2 half lives, it means that it has first reduced to its half and then again the half of what it was, i.e. 1/4</u>

Thus, after two successive half-lives, the concentration must be 1/4 of the initial concentration and hence, the statement is false.

4 0

4 years ago

Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope

morpeh [17]

<u>Answer:</u>

<u>For a:</u> The isotopic representation of iodine is $_{53}^{131}\textrm{I}$

<u>For b:</u> The isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

<u>For c:</u> The isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

<u>Explanation:</u>

The isotopic representation of an atom is: $_Z^A\textrm{X}$

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

<u>For a:</u>

We are given:

Number of neutrons = 78

Atomic number of iodine = 53 = Number of protons

Mass number = 53 + 78 = 131

Thus, the isotopic representation of iodine is $_{53}^{131}\textrm{I}$

<u>For b:</u>

We are given:

Number of neutrons = 82

Atomic number of cesium = 55 = Number of protons

Mass number = 55 + 82 = 137

Thus, the isotopic representation of cesium is $_{55}^{137}\textrm{Cs}$

<u>For c:</u>

We are given:

Number of neutrons = 52

Atomic number of strontium = 38 = Number of protons

Mass number = 38 + 52 = 90

Thus, the isotopic representation of strontium is $_{38}^{52}\textrm{Sr}$

3 0

3 years ago

Radon (Rn) is the heaviest, and only radioactive, member of Group 8A(18) (noble gases). It is a product of the disintegration of

Serggg [28]

Given data Atomic mass of Ra= 226g/mol

no. of moles =1.0/226g/mol =0.04424moles

no. of atoms in 0.044moles

no. of atoms =no. of moles x avogadro's number

= 0.044x 6.022 x10^23 = 0.264968 x 10^22

If 10^15 atoms of Ra produce 1,373*10^4 atoms of<u> Rn per second</u> then 2,66 *10^21 forms 3,658*10^10 atoms of Rn per second.

Day has 246060=86400 s

That means that 2,66x10^21 atoms of Ra produces 3,16 x10^15 atoms of Rn in a day.

N(Rn)=3.16* 10 ^15 n(Rn)=N/NA

n(Rn)=5,25*10−9 pV=nR*T

T=273.15K R=8,314

p=101325Pa V=n∗R∗T/p

V=5.25∗10^−9 ∗ 8.314 ∗ 273.15 / 101325

V=1.1810^−10 m^3 = 118 x10^-7 liters of Rn, measured at STP, are produced per day by 1.0 g of Ra

To know more about Ra here

brainly.com/question/9112754

#SPJ4

6 0

2 years ago

Stolb23 [73]

Answer:

There are 17.64% students received B+ grades.

Explanation:

It is given that,

Total number of students in chemistry class is 17

We need to find the percentage received by B+.

Number of students having B+ grades are 3 (from graph)

Required percentage = $P=\dfrac{3}{17}\times 100=17.64\%$

So, there are 17.64% students received B+ grades.

8 0

3 years ago

What is the empirical formula for CaCo3?*

GalinKa [24]

This problem is requiring the empirical formula for CaCO₃, which is its molecular formula, and turns out to be equal, this is A. CaCO3 according to the following:

<h3>Empirical formulas:</h3><h3 />

In chemistry, molecular formulas show both the actual type and number of atoms in a chemical compound, based on the elements across the periodic table and the subscripts standing for the number of atoms in the compound.

However, the empirical formula is a reduced expression of the molecular one, which shows the minimum number of atoms in a compound after simplifying to the smallest whole numbers.

In such a way, since the given compound is CaCO₃ and both Ca and C have a one as their subscript, it is not possible to simplify any further and therefore the empirical formula equals the molecular one this time, making the answer to be A. CaCO3.

Learn more about empirical formulas: brainly.com/question/1247523

5 0

2 years ago