Which statement is TRUE regarding the macroscopic and

Chemistry

1 answer:

lions [2.9K]3 months ago

7 0

Answer:

Chemists observe phenomena on a macroscopic level which informs their understanding of microscopic aspects.

Explanation:

Many critical chemical insights arise from macroscopic observations because most scientific instruments currently cannot directly evidence microscopic events. Data gathered from these larger-scale observations can yield valuable insights into the nature of specific microscopic interactions.

This is particularly true in atomic structure studies. The majority of evidence that contributed to our understanding of atomic structure was obtained from macroscopic observations and subsequently provided crucial information regarding the atom's microscopic configuration.

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Does the result of the calculation in question 3 justify your original assumption that all of the SCN^- is in the form of FeNCS^

lorasvet [2795]

Fe 3+ + SCN- --> FeSCN 2+

.......Fe 3+.......SCN-.........FeSCN 2+ 
I.......0.04..........0.001.............. 
C........-x...............-x............. 
E.....0.04-x.....0.001-x...........x 

Keq = 203.4 = x / (0.04-x)(0.001-x) 
203.4 = x / (x^2 - 0.041x + 4x10^-5) 
203.4x^2 - 8.34x + 0.00094 = x 
203.4x^2 - 9.34x + 0.00094 = 0 
x = -0.0001M or 0.0458M 
therefore, according to the calculated Keq, all of the SCN- and Fe 3+ would be fully converted into FeSCN 2+

5 0

3 months ago

A 19.3-g mixture of oxygen and argon is found to occupy a volume of 16.2 l when measured at 675.9 mmhg and 43.4oc. what is the p

KiRa [2933]

The partial pressure of oxygen is 438.0 mmHg. The ideal gas equation is expressed as PV = nRT where P represents pressure, V denotes volume, n is the number of moles, R is the ideal gas constant (8.3144598 (L*kPa)/(K*mol)), and T signifies absolute temperature. To convert from Celsius to Kelvin, we have 43.4 + 273.15 = 316.55 K. For the pressure conversion from mmHg to kPa: 675.9 mmHg * 0.133322387415 = 90.11260165 kPa. When solving for n using the ideal gas equation, we derive n = PV / (RT) which provides n = 90.11260165 kPa * 16.2 L / (8.3144598 (L*kPa)/(K*mol) * 316.55 K)= 1459.824147 L*kPa / 2631.94225 (L*kPa)/(mol), resulting in n = 0.554656603 mol. Thus, we have 0.554656603 moles of gas particles. Next, we determine the contribution from oxygen. The atomic weight of oxygen is 15.999 g/mol, while argon is 39.948 g/mol, and the molar mass of O2 is 31.998 g/mol. We establish the relationships where M is the number of moles of O2, and 0.554656603 - M gives the number of moles of Ar. Setting up the equation: M * 31.998 + (0.554656603 - M) * 39.948 = 19.3, we solve for M resulting in 0.359424148 moles of oxygen out of 0.554656603 total moles. This leads to oxygen providing 0.359424148 / 0.554656603 = 0.648012024 or 64.8012024% of the total pressure of 675.9 mmHg. The partial pressure therefore calculates to 675.9 * 0.648012024 = 437.9913271 mmHg, rounded to 438.0 mmHg

7 0

3 months ago

A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74

KiRa [2933]

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

= 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product $(CH3COOH)^{2}$ gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol $(CH3COOH)^{2}$ / 2 mol CH3COOH)

= 0.005 mol $(CH3COOH)^{2}$

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^{2}$

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

= 0.74 atm * (0.015 mol / 0.020 mol)

= 0.555 atm

4 0

3 months ago