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lidiya [134]
4 years ago
7

True or False: All temperatures used in the comparisons or calculations of gases must use the Kelvin temperature scale.

Chemistry
2 answers:
vlabodo [156]4 years ago
7 0
True. you should always use Kelvin scale for measuring gases.
Digiron [165]4 years ago
6 0

Answer:

True

Explanation:

I used my notes from class today. I could be wrong.

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When the concentration of SO2(g) is increased to 1.48 M, the ratio of products to reactants is 1.4. The equilibrium constant for
weeeeeb [17]

Answer:

Towards the products

Explanation:

3 0
3 years ago
Cl- and Ca2+
evablogger [386]

  • they are both in thier octet state

differences

  • Cl loses 1 electrons to become stable
  • Ca2 gains one electrons to become stable
7 0
3 years ago
Calculate the quantity of energy produced per gram of U-235 (atomic mass = 235.043922 amu) for the neutron-induced fission of U-
Alexus [3.1K]

Answer:

To calculate the amount of energy produced per gram from Uranium- 235 we call on the formula:

delta(m)= mass of the products - mass of reactants

Given the atomic mass of Xe-144 = 143.9385 amu; atomic mass of Sr-90= 89.907738 amu;  atomic mass of  U-235= 235.043922 amu

Therefore:

delta(m)=  (143.9385 + 89.907738) - (235.043922)  = -1.197682 amu

Recall that to calculate energy in joules, we use the formula:

Energy = mc^2

Therefore: Energy = (-1.197582/6.022 x 10^26)kg x (3 x 10^8 m/s)^2

= -1.7898104 x 10^-10 J

= (-1.7898104 x 10^-10 x 6.022 x 10^23/235.043922)

= -4.586 x 10^11 J per gram of  energy released

Explanation:

To calculate the amount of energy produced per gram from Uranium- 235 we call on the formula:

delta(m)= mass of the products - mass of reactants

Given the atomic mass of Xe-144 = 143.9385 amu; atomic mass of Sr-90= 89.907738 amu;  atomic mass of  U-235= 235.043922 amu

Therefore:

delta(m)=  (143.9385 + 89.907738) - (235.043922)  = -1.197682 amu

Recall that to calculate energy in joules, we use the formula:

Energy = mc^2

Therefore: Energy = (-1.197582/6.022 x 10^26)kg x (3 x 10^8 m/s)^2

= -1.7898104 x 10^-10 J

= (-1.7898104 x 10^-10 x 6.022 x 10^23/235.043922)

= -4.586 x 10^11 J per gram of  energy released

7 0
4 years ago
Elements listed in vertical columns of the periodic table:
marysya [2.9K]
Vertical columns are called groups, or families. 

They share similar chemical properties. 

Answer: <span>have similar chemical properties</span>
3 0
3 years ago
Read 2 more answers
Calculate the energy change when an electron moves from n=5 to n=7. Explain/show work please.
Korolek [52]

Answer: E = 1.55 ⋅ 10 − 19 J

Explanation:  

The energy transition will be equal to  1.55 ⋅ 10 − 1 J .  

So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

1 λ  = R ⋅ ( 1 n 2 final  − 1 n 2 initial  ) , where λ - the wavelength of the emitted photon; R

- Rydberg's constant -  1.0974 ⋅ 10 7 m − 1 ; n final - the final energy level - in your case equal to 3; n initial - the initial energy level - in your case equal to 5. So, you've got all you need to solve for  λ , so 1 λ  =

1.0974 ⋅10  7 m − 1 ⋅ (....     −152    

)

1

λ

=

0.07804

⋅

10

7

m

−

1

⇒

λ

=

1.28

⋅

10

−

6

m

Since  

E

=

h

c

λ

, to calculate for the energy of this transition you'll have to multiply Rydberg's equation by  

h

⋅

c

, where

h

- Planck's constant -  

6.626

⋅

10

−

34

J

⋅

s

c

- the speed of light -  

299,792,458 m/s

So, the transition energy for your particular transition (which is part of the Paschen Series) is  

E

=

6.626

⋅

10

−

34

J

⋅

s

⋅

299,792,458

m/s

1.28

⋅

10

−

6

m

E

=

1.55

⋅

10

−

19

J

8 0
3 years ago
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