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3 years ago

The sodium atom loses an electron to form a sodium ion (Na+). Which statement is correct with respect to its atomic radius?

2 answers:

6 0

The atomic number of sodium is 11. Sodium is a metal and it has the electronic configuration: 2, 8,1. This configuration implies that, the electrons in an atom of sodium are distributed into three different shells and the outermost shell has one electron in it. If sodium decide to give away the electron on its outermost shell, it will have 10 electrons left and those electrons will be distributed in only the first two shells and then it will be described as an ion. The third shell will not exist again because the electron there has been given away. Thus, the sodium ion is going to have a smaller atomic radius because its size has been reduced. This implies that the sodium ion will have a smaller radius than the sodium atom.

mixer [17]3 years ago

3 0

Its atomic radius increases because there is less electron shielding after the sodium loses one e-. Sodium is a nonmetal which means it forms an anion. Anions always increase from their original atomic radius

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Zinc dissolves in hydrochloric acid to yield hydrogen gas: Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g) When a 12.7 g chunk of zinc

ollegr [7]

Answer:

$\boxed{\text{0.673 mol/L}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Data:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 65.38

Zn + 2HCl ⟶ ZnCl₂ + H₂

m/g: 12.7

V/mL: 5.00×10²

c/mol·L⁻¹: 1.450

2. Moles of each reactant

(a) Moles of Zn

$n = \text{12.7 g Zn} \times \dfrac{\text{1 mol Zn}}{\text{65.38 g Zn}} = \text{0.1942 mol Zn}$

(b) Moles of HCl

V = 5.0× 10² mL = 0.5000 L

$n = \text{0.5000 L HCl}\times \dfrac{\text{1.450 mol HCl}}{\text{1 L HCl}} = \text{0.7250 mol HCl}$

3. Identify the limiting reactant

Calculate the moles of ZnCl₂ obtained from each reactant

(i) From Zn

The molar ratio is 1 mol ZnCl₂:1 mol Zn

$n = \text{0.1942 mol Zn} \times \dfrac{\text{1 mol ZnCl}_{2}}{\text{1 mol Zn}} = \text{0.1942 mol ZnCl}_{2}$

(ii) From HCl

The molar ratio is 1 mol ZnCl₂:2 mol HCl

$n = \text{0.7250 mol HCl} \times \dfrac{\text{1 mol ZnCl}_{2}}{\text{2 mol HCl}} = \text{0.3625 mol ZnCl}_{2}$

Zinc is the limiting reactant, because it produces fewer moles of ZnCl₂.

4. Moles of HCl reacted

The molar ratio is 2 mol HCl:1 mol Zn

$n = \text{0.1942 mol Zn} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Zn}} = \boxed{\text{0.3885 mol HCl}}$

5. Moles of HCl remaining

n = 0.7250 - 0.3885 = 0.3365 mol HCl

6. Concentration of hydrogen ions

The HCl is completely dissociated.

$c = \dfrac{\text{0.3365 mol}}{\text{0.5000 L}} = \textbf{0.673 mol/L}\\\\\text{The concentration of hydrogen ions is $\boxed{\textbf{0.673 mol/L}}$}$

4 0

3 years ago

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.

timofeeve [1]

Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU + w

q = 475 J + 93 J

q = 568 J

Therefore, the magnitude of q for the process 568 J.

6 0

3 years ago

PLEASE HELP!!!!! I'LL YOU BRAINIEST!!!!

Sidana [21]

Answer:

I believe 11 is B 12 is C 13 is B and 14 is C

3 0

2 years ago

What will be the final temperature of the solution in a coffee cup calorimeter if a 50.00 mL sample of 0.250 M HCl(aq) is added

padilas [110]

Answer:

21.21°C will be the final temperature of the solution in a coffee cup calorimeter.

Explanation:

$HCl+NaOH\rightarrow H_2O+NaCl$

$\Delta H$ = enthalpy change = -57.2 kJ/mol of NaOH

Moles of sodium hydroxide = n

Molarity of the NaOH = 0.250 M

Volume of NaOH solution = V = 50.00 mL = 0.050 L

$n=Molarity\times V=0.250 M\times 0.050 L= 0.0125 mol$

Moles of HCl = n'

Molarity of the HCl= 0.250 M

Volume of HCl solution = V' = 50.00 mL = 0.050 L

$n'=Molarity\times V=0.250 M\times 0.050 L= 0.0125 mol$

Since 1 mole of Hcl reacts with 1 mole of NaoH. Then 0.0125 mole of HCl will react with 0.0125 mole of NaOH.

The enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

$q=\Delta H\times n=-57.2 kJ/mol \times 0.0125 mol= -0.715 kJ=-715 J$

q = heat released on reaction= -715 J

now, we calculate the heat gained by the solution.:

Q= -q = -(-715 J) = 715 J

m = mass of the solution = ?

Volume of the solution formed by mixing, v = 50.00 ml + 50.00 mL = 100.00 mL

Density of the solution = density of water = d = 1 g/mL

$mass=density\times volume=d\times v=1 g/ml \times 100.00 ml=100 g$

m = 100 g

q = heat gained = ?

c = specific heat = $4.18 J/^oC$

$T_{f}$ = final temperature = ?

$T_{i}$ = initial temperature = $19.50^oC$

$Q=mc\times (T_{f}-T_{i})$

Now put all the given values in the above formula, we get:

$715 J=100 g\times 4.18J/^oC\times (T_f-19.50)^oC$

$T_f=21.21 ^oC$

21.21°C will be the final temperature of the solution in a coffee cup calorimeter.

5 0

3 years ago

How does the energy of an electron change when the electron moves closer to the nucleus

Minchanka [31]

When a electron moves from higher shell to lower shell ,it releases energy because if it goes to lower shell without releasing energy then it will unable to do that so.as lower electron shells have low energy electrons

8 0

3 years ago