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Elis [28]
4 years ago
14

Without the ______, the greenhouse effect would not be possible.

Chemistry
1 answer:
SVETLANKA909090 [29]4 years ago
4 0
Without the atmosphere, the greenhouse effect would not be possible.
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Which of the following is the BEST description of a temperate grassland ecosystem?
Likurg_2 [28]

Answer:

It is an area that is covered in grasses and wildflowers that receives enough rainfall to support the grassland but not the growth of trees.

Explanation:

8 0
2 years ago
A bottle in lab is labeled [CoCl2.6H2O] = 0.71 M in 9.217 M HCl. If you determine [CoCl42-] to be 0.361 M at a particular temper
77julia77 [94]

Answer:

[Co(H₂O)₆]²⁺ = 0.361 M

Explanation:

[CoCl₂ . 6H₂O] + 2Cl⁻ → [CoCl₄]²⁻ + 6H₂O

[Co(H₂O)₆]²⁺ + 4Cl⁻ → [CoCl₄]²⁻ + 6H₂O

[CoCl₄]²⁻ = 0.361 M ∴ [Co(H₂O)₆]²⁺ = 0.361 M

The equation shows that the concentration of [Co(H₂O)₆]²⁺ should be equal to the concentration of [CoCl₄]²⁻.

3 0
4 years ago
Please help ASAP!! I’m stuck!!
Gnoma [55]

Answer:

The answer of this question is 3

8 0
3 years ago
Read 2 more answers
Which of the following tells you the number of molecules in 1 mole of a gas?
Nezavi [6.7K]

Answer:

Avogrado's number

Explanation:

It's unniversal measure of moles

7 0
3 years ago
Nitrogen gas at 300 k and 200 kpa is throttled adiabatically to a pressure of 100 kpa if the change in kinetic energy is negligb
erastovalidia [21]

Answer:

The temperature of the Nitrogen after throttling is  T_2 = 300 \  K

Explanation:

From the question we are told that

   The  temperature is  T_1 =  300 \  K

    The pressure is  P =  200 \  kPa  =  200 * 10^{3} \  Pa

    The pressure after being  P_1 =  100 \  kPa =  100 * 10^{3} \  Pa

   

Generally from the first law of thermodynamics we have that

      Q - W =  \Delta U + \Delta  K

Here \Delta U  is the change internal  energy which is mathematically represented as

            \Delta U =  C_p  (T_2 - T_1)

Here      C_p is the specific heat of the gas at constant pressure

        \Delta K  is the change kinetic energy which is negligible

         Q  is the thermal  energy  which is Zero for an adiabatic process

         W  is the work done and the value is zero given that the gas was throttled adiabatically

So

      0=  \Delta U +0

=>   \Delta U  = 0

=>  (T_2 - 300)  = 0

=>   T_2 = 300 \  K

     

3 0
3 years ago
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