Which element has the electron configuration 2-4?

3. Write DISSOCIATION or IONIZATION equations for the following chemicals:

Romashka [77]

Answer:

a) HNO3 -> H+ + NO3- disassociation of Nitric Acid; to yield a Nitrate ion and a Proton, H+, or as a Hydronium ion H3O+

b) H2S04 -> Disassociation of Sulfuric Acid; simple way- 2H+ + SO4- -

c) H2S hydrogen sulphide in water is an acid; thus H+ HS- disassociation.

d) NaOH -> dissociation of Na+ + OH-; this is complete; sodium hydroxide is deliquescent, meaning it will draw water - EVEN from the air! Strong Base

e) Na2CO3 -> 2Na+ CO3- - Ionization of sodium carbonate - a salt

f) Na2S04 -> 2Na+ + SO4 - - ionization of sodium sulphate - a salt

g) NaCl -> Na+ + Cl- ionization of the salt, Sodium Chloride

Explanation:

Salts ionize at different rates; acids or bases dissociate; these are mostly strong acids and NaOH, a strong base.

8 0

2 years ago

A 55.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a

balu736 [363]

Answer:

0.14 M

Explanation:

To determinate the concentration of a new solution, we can use the equation below:

C1xV1 = C2xV2

Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL

1.50x55.0 = C2x278

C2 = 0.30 M

The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL

Then,

0.30x139 = C2x294

C2 = 0.14 M

4 0

2 years ago

Read 2 more answers

A bomb calorimeter has a heat capacity of 783 J/oC and contains 254 g of water whose specific heat capacity is 4.184 J/goC. How

IrinaK [193]

Answer : The amount of heat evolved by a reaction is, 4.81 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $783J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 254 g

$\Delta T$ = change in temperature = $T_2-T_1=(23.73-26.01)=-2.28^oC$

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

$q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]$

$q=-4208.28J=-4.81kJ$

Therefore, the amount of heat evolved by a reaction is, 4.81 kJ

7 0

3 years ago

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell = 0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0

3 years ago

Does a large or small body cool faster?

Nataly [62]

Answer:

SO… The larger wire looses heat energy faster, however the smaller wire decreases temperature faster. ... Their surface area is much larger in proportion to their body mass and they lose heat through their skin when it is cold and they gain heat through their skin when it is hot much faster than an adult does.

Explanation:

8 0

2 years ago