Please help!! Why is the following Electron Configuration incorrect for Aluminium?

A student measures the mass of a 6.0 cm3 block of brown sugar to be 10.0 g. What is the density of the brown sugar?

Alex17521 [72]

Answer:

1.67g/cm3

Explanation:

The formula for density is $d=\frac{m}{v}$ . The m variable stands for mass and the v variable stands for volume.

The mass of the brown sugar is 10.0g and the volume is 6.0cm3, so we can plug those values into the equation.

$d=\frac{10g}{6cm^{3} }$

$d=1.67$ $\frac{g}{cm^{3} }$

Rounded to 3 significant figures, the density of the block of brown sugar is 1.67 g/cm3. If the mass is in grams and the volume is in cm3, the unit for the final answer is $\frac{g}{cm^{3} }$ (grams per centimetres cubed).

8 0

3 years ago

Read 2 more answers

The question is below

rodikova [14]

The anode is the electrode where the oxidation occurs.

Cathode is the electrode where the reducction occurs.

Equations:

Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V

The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.

Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72

Answer: 2.72 V

3 0

3 years ago

How many PO43? ions are in a mole of K3PO4?

adoni [48]

K₃PO₄ → 3K⁺ (aq) + PO₄³⁻(aq)

One mole of PO₄³⁻ ion gets dissociated from one mole of K₃PO₄

As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ ions

One mole of PO₄³⁻ ions x (6.022 x 10²³ ions/ 1 mole of PO₄³⁻ ions )

= 6.022 x 10²³ ions

Therefore , there are 6.022 x 10²³ PO₄³⁻ ions in a mole of K₃PO₄.

7 0

3 years ago

The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi

avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

$Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)$

The balanced two-half reactions will be,

Oxidation half reaction : $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction : $Ni^{2+}+2e^-\rightarrow Ni$

The expression for reaction quotient will be :

$Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

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

$Q=\frac{(0.0141)}{(0.00104)}=13.6$

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

$E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q$

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

$E^o_{cell}$ = standard electrode potential of the cell = 0.51 V

$E_{cell}$ = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

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

$E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)$

$E_{cell}=0.47V$

Therefore, the actual cell potential of the cell is 0.47 V

4 0

3 years ago

The ionization energies for removing successive electrons from sodium are 496 kJ/mol, 4562 kJ/mol,

ddd [48]

Answer:

D

Explanation:

The high jump of ionization energy indicates that we are trying to remove electron from noble gas configuration state.

The ionization energy data specifies that the Elements are from group 1 at period 3 or greater.

Removing the first electron require 496 kJ and the second ionization energy jump significantly due to the removal of electron from the noble gas configuration which is logical because electron try to maintain the especially stable state.

3 0

3 years ago