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Dmitrij [34]
2 years ago
11

1 what do you understand by the term isotopes 2 why do I should talk of an element process identical chemical properties 3 name

the isotopes of Hydrogen which does not have a Newton in it​
Chemistry
1 answer:
Bad White [126]2 years ago
5 0

Answer:

Isotopes of an element have same number of protons but different number of neutrons. Which means isotopes of an element have same atomic number but different mass number.

The chemical property of an element is determined by the number of electrons. And as all the isotopes have same number of electrons, they have same chemical properties.

Thus as isotopes of an element have same atomic number , they have same number of electrons and protons. As they have different mass number, the number of neutrons will be different. Hydrogen has three isotopes , ^1_1\textrm{H}, ^2_1\textrm{H} and ^3_1\textrm{H}. Thus ^1_1\textrm{H} has no neutron.

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Ede4ka [16]

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what is scientific model

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4 0
2 years ago
In molecules, energy is stored in?<br> Atoms<br> Electrons<br> The Nucleus<br> Bonds
Nikolay [14]

Answer:

Atoms

Explanation:

Energy, potential energy, is stored in the covalent bonds holding atoms together in the form of molecules. This is often called chemical energy.

4 0
3 years ago
Difference between emprical and molecular formula
Musya8 [376]

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3 0
2 years ago
Given that the molar mass of NaNO3 is 85.00 g/mol, what mass of NaNO3 is needed to make 4.50 L of a 1.50 M NaNO3 solution?
marissa [1.9K]
Mass = molarity x molar mass( NaNO₃) x volume

mass = 1.50 x 85.00 x 4.50

mass = 573.75 g of NaNO₃

hope this helps!
8 0
3 years ago
Read 2 more answers
A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
RoseWind [281]

Answer:

C_3=0.125M

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

V_1C_1=V_2C_2

So we solve for C2:

C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M

2. Now, since 111 mL of water is added, we compute the final volume, V3:

V_3=139+111=250mL

So, the final concentration of the 139 mL portion is:

C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M

Best regards!

8 0
3 years ago
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