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inna [77]
3 years ago
7

For the following questions, use a periodic table and your atomic calculations to find the unknown information about each isotop

e:
You have a Helium Isotope with 2 neutrons. What is the mass number?

2
4
4.0026
6
Chemistry
1 answer:
mafiozo [28]3 years ago
7 0

Answer:

Option B is correct.

4

Explanation:

We know that an atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example, if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.

In given problem we are given with 2 neutrons of helium. We know that the atomic number of He is 2. Thus Mass number of He is,

Number of neutrons + number of proton

2 + 2 = 4

Thus, option B is correct.

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Answer:

4.occupy a definite volume while taking the shape of their container.

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8 0
3 years ago
write the chemical formula of the following compounds sodium sulfate magnesium chloride aluminum nitrate ammonium sulfate​
tekilochka [14]

Explanation:

Magnesium chloride = MgCl₂

aluminum nitrate= Al(NO₃)₃

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3 years ago
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OlgaM077 [116]

Answer:

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Explanation:

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8 0
3 years ago
Suppose a student titrates a 10.00-ml aliquot of saturated ca(oh)2 solution to the equivalence point with 16.08 ml of 0.0199 m h
Alborosie
Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O

Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O

Given,
concentration of HCl = 0.0199 M
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∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
                                                           = 0.0199 X 16.08 X 10^(-3)
                                                           = 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = 
\frac{number of moles of [OH-]}{volume of solution (L)} = \frac{3.1999 X 10^(^-^4^)}{10 X 10^(^-^3^)} = 0.03199 M
6 0
3 years ago
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