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pickupchik [31]
3 years ago
9

The four sets of lines in the hydrogen emission spectrum are known as Balmer, Brackett, Paschen, and Lyman. For each series, ass

ign the energy (infrared, ultraviolet, or visible), the nr value, and all possible n, values up to 7 Series Balmer Brackett Paschen Lyman Energy nf ni visible) infrared ultraviolet 1 2 3 6 7

Chemistry
1 answer:
arlik [135]3 years ago
8 0

Answer:

Hydrogen spectrum

Explanation:

Balmer series - Observed in the visible region

Brackett series - Observed in the infrared region

Paschen series - Observed in the infrared region

Lyman series - Observe in the Ultraviolet region.

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How to balance the question?​
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Answer:

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CH4 + 2O2

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So ; 1 C , 4 H , 4 O

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Cell Membranes
Lerok [7]

The procedure to perform a cell membrane experiment is to:

  • Use beetroots cells to measure the permeability of the membrane
  • This is done to check the content of the pigment
  • It is also meant to check the pigment leaks out of the cells.

<h3>What is a Cell Membrane?</h3>

This refers to the semi-permeable membrane that is around the cytoplasm of a cell.

Hence, we can see that the main purpose of the cell membrane practical experiment is to test the permeability of a membrane and to see the amount of liquid that a membrane can hold.

Please note that your question is incomplete so I gave you a general overview to get a better understanding of the concept.

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brainly.com/question/1768729

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4 0
2 years ago
What mass of water could be warmed from 21.4 degrees celsius to 43.4 degrees celsius by the pellet dropped inside it? Heat capac
Artist 52 [7]

42.34 g of water could be warmed from 21.4°C to 43.4°C  by the pellet dropped inside it

Heat loss by the pellet is equal to the Heat gained by the water.

q_{w} = -q_{p} ….(1)

where, q_{w} is the heat gained by water

q_{p} is the heat loss by pellet

q_{w} = mCΔT

where m = mass of water

C = specific heat capacity of water = 4.184 J/g-°C

ΔT = Increase in temperature

ΔT for water = 43.4 - 21.4 = 22°C

q_{w} = m × 4.184 × 22 …. (2)

Now

q_{p} = H_{c} ×ΔT

where H_{c} = Heat capacity of pellet = 56J/°C

Δ T for pellet = 43.4 - 113 =- 69.6°C

q_{p} = 56 × -69.6 = -3897.6 J

From equation (1) and (2)

-m× 4.184 × 22 =-3897.6

m= 42.34 g

Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.

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6 0
1 year ago
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