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Andrew [12]
3 years ago
6

Definition of isotopes

Chemistry
2 answers:
sukhopar [10]3 years ago
5 0
Isotopes are atoms of the same element that have different numbers of neutrons.
klasskru [66]3 years ago
3 0
Isotopes are element that exhibits isotopy ie they have the same atomic no but different mass on
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The Lewis dot notation for two atoms is shown.
Iteru [2.4K]

Answer:

Mg donates two electrons to O

Explanation:

Lewis dot notation uses dots and crosses to represent valence electrons on atoms.

Magnesium is a metal and would donate or lose electrons during bonding.

Oxygen is a non metal and would gain electrons during bonding.

The correct option is;

Mg donates two electrons to O

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2 years ago
Many gyms and health clubs have steam saunas, which are small steam-filled rooms. Traditionally, steam saunas have a container o
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3 years ago
Which word equation represents a neutralization reaction
Vlad1618 [11]
Base+salt > acid +alkali > neutralization i think this is the reaction
8 0
2 years ago
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26.0 g of a liquid that has a density of 1.44 g/mL needs to be measured out in a graduated cylinder. what volume should be used
Vaselesa [24]
As Density = Mass/Volume
Mass = 26.0g 
Density = 1.44g/mL
Therefore Volume = Mass/Density
=> Volume = 26.0/1.44 = 18.055... = 18.1mL (to 3 sig figs)

4 0
2 years ago
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A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of
jek_recluse [69]

<u>Answer:</u> The volume when the pressure and temperature has changed is 1.6\times 10^2mL

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

Let us assume:

P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K

Putting values in above equation, we get:

\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL

Hence, the volume when the pressure and temperature has changed is 1.6\times 10^2mL

5 0
3 years ago
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