What happens when an electron jumps from energy level 1 to energy level 2 in an atom?

Zippy decides he needs a drink to wash down the cold Pring Pongs. There are two 355 mL cans of Kookie Cola in the mini-bar.

Lady bird [3.3K]

20.7 Degree celsius is the temperature of Zippy's room.

Explanation:

The initial conditions of the kookie cola will be depicted by P1,V1 and T1.

P1= 36.5 psi or 2.4816 atm

V1= 355ml or 0.35 L

T1= 275.5K

The final conditions are given P2,V2 and T2

P2= 38.9 psi or 2.6469 atm

V2 = 355 ml or 0.35 L

also the solubility of the carbon dioxide does not change with temperature.

Applying the formula

P1VI/T1=P2V2/T2

since volume does not change and remains constant.

It can be written as:

P1/T1=P2/T2

= 2.4816/275.5=2.6469/T2

T2= 293.85 K

7 0

3 years ago

Is na=na+ reduced or oxidized

Sonbull [250]

Answer:

oxidized

Explanation:

Each sodium atom loses an electron to form a sodium ion.

8 0

3 years ago

2.

Alex777 [14]

Answer:

C. 500 cm' of 1.0 mol dmº magnesium sulphate solution.

Explanation:

Let us look at each of the solutions individually;

CaCl2 has three particles

K2SO4 has three particles

MgSO4 has two particles

C2H5OH has only one particle

The number of moles of moles in 250 cm of 2.0 mol dm-3 potassium chloride is 250/1000 * 2 = 0.5 moles having two particles

Also; number of moles in 500 cm' of 1.0 mol dm-3 magnesium sulphate solution= 500/1000 * 1 = 0.5 moles having two particles

5 0

3 years ago

When solid ammonium carbamate sublimes, it dissociates completely into ammonia and carbon dioxide according to the following equ

Harlamova29_29 [7]

Answer: The equilibrium constant for the given equation is 0.0315

Explanation:

We are given:

Total pressure in the container = 0.596 atm

The given chemical equation follows:

$N_2H_6CO_2(s)\rightleftharpoons 2NH_3(g)+CO_2(g)$

Initial: -

At eqllm: - 2x x

Evaluating the value of 'x'

$\Rightarrow (x+2x)=0.596\\\\x=0.199$

The expression of $K_p$ for above equation follows:

$K_p=p_{CO_2}\times (p_{NH_3})^2$

The partial pressure of pure solids and pure liquids are taken as 1 in the equilibrium constant expression.

Putting values in above expression, we get:

$K_p=(0.199)\times (2\times 0.199)^2\\\\K_p=0.0315$

Hence, the equilibrium constant for the given equation is 0.0315

7 0

3 years ago

EQUILIBRIUM Two moles of compound P were placed in a vessel. The vessel was heated and compound P was partially decomposed to pr

Reika [66]

As given:

Initial moles of P taken = 2 mol

the products are R and Q

at equilibrium the moles of

R = x

total moles = 2 + x/2

Let us check for each reaction

A) P <-> 2Q+R

Here if x moles of P gets decomposed it will give 2x moles of Q and x moles of R

So at equilibrium

moles of P left = 2- x

moles of Q = 2x

moles of R = x

Total moles = (2-x) + 2x + x = 2 +2x

B) 2P <-> 2Q+R

Here x moles of P will give x moles of Q and x/2 moles of R

So at equilibrium

moles of P left = 2- x

moles of Q = x

moles of R = x/2

Total moles = (2-x) + x + x/2 = 2 + x/2

C) 2P <-> Q+R

Here x moles of P will give x/2 moles of Q and x/2 moles of R

So at equilibrium

moles of P left = 2- x

moles of Q = x /2

moles of R = x/2

Total moles = (2-x) + x + x = 2

D) 2P <-> Q+2R

Here x moles of P will give x/2 moles of Q and x moles of R

So at equilibrium

moles of P left = 2-x

moles of Q = x/2

moles of R = x

Total moles = (2-x) + x/2 + x = 2 + x/2

3 0

3 years ago