All categories

3 years ago

Use the information below to explain why the atomic radius decreases down a group.

Chemistry

1 answer:

notsponge [240]3 years ago

3 0

Answer:

Detail is given below

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm

In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N

In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.

You might be interested in

Please answer this question for 10 points

Dmitry_Shevchenko [17]

Is A have a nice day good luck b

4 0

3 years ago

Read 2 more answers

As the temperature of one of the four liquids increases, the vapor pressure will

s2008m [1.1K]

Answer:

a) increase exponentially.

Explanation:

The vapor pressure is depend only on temperature.

The vapor pressure of liquid does not depend upon amount of liquid. For example whether the liquid is 50 g or 30 g its vapor pressure will remain same according to the temperature.

The temperature and vapor pressure have exponential relationship. As the temperature of liquid increases its vapor pressure also goes to increase. When the temperature of liquid goes to decrease its vapor pressure also decreases.

The change in vapor pressure of substance when temperature changes is given as,

ln P₂/P₁ = ΔH(va)/R (1/T₁ - 1/T₂)

7 0

3 years ago

Alcl3+Na =NaCl+ Al .did Al change oxidation number

saul85 [17]

Answer:

yes it is ( From +3 to 0 )

Explanation:

If this is the balanced equation:

AlCl3 + 3Na ——> 3NaCl + Al

Al Cl 3Na Na Cl Al

+3 -3 0 +1 -1 0

5 0

3 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.

Back to this question. $[\text{H}^{+}]$ is given. 25 °C implies that $K_w = 10^{-14}$ . As a result,

$\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

Andrews [41]

sorry what us this i don't get it

7 0

2 years ago