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djyliett [7]
3 years ago
10

Can anyone do these? or even just help .. It is urgent and I've put up all of my points. I've been able to complete everything e

lse but this and I'm stuck.

Chemistry
1 answer:
Lorico [155]3 years ago
7 0
A3B2
bond is ionic 
A is in group 2 (you can pick any like Ca)
B is in group 5 (like B)
the other question:
the reason is they are neutral gas and they already have 8 electrons except for He which is 2 and are completely stable so don't want to loose any electron vs Li and Na which have only 1 electron in the outer layer and are willing to loose that one to become stable. 


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Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc
wolverine [178]

Answer:

(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔2Na[Al(OH)4](aq) + 3H2(g)

∴ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11

(2) H2O(l) + SO3(g) ↔ H2SO4(aq)

∴ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)

∴ Kc = Kc = 1 / PO2∧6

Explanation:

(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔ 2Na[Al(OH)4](aq) + 3H2(g)

∴ O /  Al: 0 → +2 ≡ 2e-

         Na: +1 → +2

∴ R /  H: +1 → 0

     2 - Al - 2

     2 - Na - 1

     8 - O - 8

     14 - H - 14

⇒ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11

(2) H2O(l) + SO3(g) ↔ H2SO4(aq)

           1 - S - 1

           4 - O - 4

           2 - H - 2

⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123

(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)

       8 - P - 8

       12 - O - 12

⇒ Kc = 1 / PO2∧6

6 0
3 years ago
5 examples of Earth Surface .<br>​
Solnce55 [7]

Answer:

<em>lithosphere</em><em> </em><em>(land)</em>

<em>hydrosphere (water)</em>

<em>biosphere</em><em> </em><em>(living things)</em>

<em>atmosphere (air).</em>

6 0
2 years ago
Read 2 more answers
The unheated Gas in the above system has a volume of 20.0 L at a temperature of 25.0 C and a pressure of 1.00 atm. The gas is he
kipiarov [429]

Answer:

1.25 atm.

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Initial volume (V1) = 20L

Initial temperature (T1) = 25°C

Initial pressure = 1 atm

Final temperature (T2) = 100°C

Final volume (V2) = constant i.e remain the same

Final pressure (P2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

Temperature (Kelvin) = temperature (celsius) + 273

Initial temperature (T1) = 25°C = 25°C + 273 = 298K

Final temperature (T2) = 100°C = 100°C + 273 = 373K

Step 3:

Determination of the final pressure of the gas. This is illustrated below:

Since the volume is constant, the following equation, P1/T1 = P2/T2 will be used to obtain the final pressure of gas as follow:

P1/T1 = P2/T2

Initial temperature (T1) = 298k

Initial pressure = 1 atm

Final temperature (T2) = 373K

Final pressure (P2) =?

P1/T1 = P2/T2

1/298 = P2 /373

Cross multiply to express in linear form

298 x P2 = 1 x 373

Divide both side by 298

P2 = 373/298

P2 = 1.25 atm.

Therefore, the pressure of the heated gas is 1.25 atm.

3 0
3 years ago
A container of gas is heated from 250 K to 303 K. What is the new pressure if the initial pressure is 880 kPa? (Assume constant
evablogger [386]

Answer:

The new pressure at constant volume is 1066.56 kPa

Explanation:

Assuming constant volume, the pressure is diectly proportional to the temperature of a gas.

Mathematically, P1/T1 = P2 /T2

P1 = 880 kPA= 880 *10^3 Pa

T1 = 250 K

T2 = 303 K

P2 =?

Substituting for P2

P2 = P1 T2/ T1

P2 = 880 kPa * 303 / 250

P2 = 266,640 kPa/ 250

P2 = 1066.56 kPa.

The new pressure of the gas is 1066.56 kPa

5 0
3 years ago
What mass, in grams, of bromine gas (Br2) is contained in a 15.7-liter tank at 24.6 degrees Celsius and 0.986 atmospheres? Show
Nezavi [6.7K]

PV = nRT

P is pressure, V is volume, n is number of moles, R is the gas constant, T is temperature in K

(2.85 atm)(12.5 L) = (n)(.08206)(27 C + 273)

n = 1.45 moles x 35.45 grams / mol Cl2 = 51.3 grams

4 0
3 years ago
Read 2 more answers
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