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AURORKA [14]
3 years ago
10

Choose a reasonable explanation to account for the differences. There may be more than one possible reason that makes sense, jus

t select one of them.
A. It is possible not all of the water was evaporated from the sand, causing the recovered mass to be higher.
B. It is possible not all of the water was evaporated from the sand, causing the recovered mass to be lower.
C. While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the recovered mass to be lower
D. While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the recovered mass to be higher.
E. There was no difference in recovered and original mass, so there is no difference to account for.
Chemistry
1 answer:
Ugo [173]3 years ago
6 0

Answer:

A. It is possible not all of the water was evaporated from the sand, causing the recovered mass to be higher

D. While drying the NaCl, the liquid boiled and some splattered out of the evaporating dish, causing the recovered mass to be higher.

Explanation:

Sand absorbs water and stores it. The sunlight causes the water to evaporate but sand can hold some of the water inside it. This results in increase in mass of the sand. The mass of sand before and after the water evaporation can be different.

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Answer:

Helium and Neon

Explanation:

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The brick is solid because it's not hollow or containing spaces or gaps
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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

⇒ <em>C</em> NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

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+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [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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Answer:

a) ΔGrxn = 6.7 kJ/mol

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Explanation:

a) The reaction is:

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The expression for Gibbs energy is:

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Where

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T = 298 K

R = 8.314 J/mol K

lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066

c) The equilibrium pressure of O₂ over M is:

K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm

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