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3 years ago

What is electron shielding

2 answers:

8 0

Shielding electrons are the electrons in the energy levels between the nucleus and the valence electrons. They are called "shielding" electrons because they "shield" the valence electrons from the force of attraction exerted by the positive charge in the nucleus. Hope this helps!!

Andreyy893 years ago

3 0

Describes the attraction between an electron and the nucleus in any atom with more than one electron

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wariber [46]

The relationship between independent and dependent variables is constant. The independent variables are not affected and do not change while dependent variables rely on something or a particular change.

Hope this helps and have a good day/night!

5 0

3 years ago

The normal freezing point of a certain liquid

stepladder [879]

Answer : The molal freezing point depression constant of X is $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of X liquid (solvent) = 450.0 g

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X= $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant of X = ?

m = molality

Now put all the given values in this formula, we get

$[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}$

$k_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of X is $4.12^oC/m$

4 0

3 years ago

A solution of H2SO4(aq) with a molal concentration of 4.80 m has a density of 1.249 g/mL. What is the molar concentration of thi

Naddika [18.5K]

Imagine we have <span>mass of solvent 1kg (1000g)
According to that: </span> $molality =n(solute) / m(solvent) =\ \textgreater \ 4.8m = 4.8mole / 1kg$
= 4.8 mole * 98 g/mole = 470g
$m(H2SO4) = n(H2SO4)*Mr(H2SO4) =\ \textgreater \$ $=\ \textgreater \ Molarity = 4.8 mole / 1.2 L = 4 M$ m(H2SO4) which is =<span>470g

</span><span>m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =></span>
$Molarity = n(solute) / V(solution) =\ \textgreater \$

5 0

3 years ago

Calculate the mass of NaCO3 used in experiment. SHOW WORK — 15 points!!

Liono4ka [1.6K]

The mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g

<h3>Calculating mass </h3>

From the question we are to calculate the mass of NaHCO₃ (sodium bicarbonate) used in the experiment

From the given information

Mass of empty evaporating dish = 46.233g

Mass of evaporating dish + Sodium bicarbonate = 48.230g

∴ Mass of sodium bicarbonate (NaHCO₃) = [Mass of evaporating dish + Sodium bicarbonate] - [Mass of empty evaporating dish]

Mass of sodium bicarbonate (NaHCO₃) = 48.230g - 46.233g

Mass of sodium bicarbonate (NaHCO₃) = 1.997 g

Hence, the mass of sodium bicarbonate (NaHCO₃) used in the experiment is 1.997 g

Learn more on Calculating mass here: brainly.com/question/15268826

5 0

2 years ago

How do you solve for the Atomic Mass in chemistry?

loris [4]

Https://www.google.com/search?q=how+to+solve+fir+atomic+mass+in+chemisty&ie=UTF-8&oe=UTF-8&hl=en-us&client=safari#kpvalbx=1
Here is the link to a great video that explains your question nicely, hope this helps.

6 0

3 years ago

Read 2 more answers