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GREYUIT [131]
3 years ago
13

Why are the elements 57 through 71 and 89 through 102, found separately at the bottom of the table?

Chemistry
1 answer:
Dovator [93]3 years ago
8 0
Elements from 57 to 71 are Lanthanoids and from 89 to 103 are Actinoids. Simply I would say Actinoids have less similarity in chemical properties than Lanthanoids. On the other hands, Actinoids have randomized chemical strutural bonds.
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Which is a TRUE statement about salt water?
makkiz [27]
I think the best answer from the given choices is option D. <span>Salt water is made up of more than one pure substance. It consists of salt and water. When each component exist alone in a container they are considered as pure substance however when mixed they are now a solution.</span>
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3 years ago
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A 10.0 g ice cube is placed into 250 g of water with an initial temperature of 20.0 C. If the water drops to a temperature of 16
Radda [10]

the mass of ice taken = 10 g

the mass of water = 250 g

initial temperature of water = 20 C

the final temperature of water = 16. 8 C

specific heat of water = 4.18 J/g*K

the heat absorbed by ice to melt = heat loss by water

heat loss by water = mass X specific heat of water X change in temperature

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heat gained by ice = 3344 J

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3 0
3 years ago
If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
vodomira [7]

Answer:

pH = 4.543

Explanation:

  • CH3CH2COOH  + H2O ↔ CH3CH2COO-  +  H3O+
  • pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

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mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

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⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

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3 years ago
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