What you understand by macrocylic effect in coordination chemistry

Answers

Answer 1

Answer:

Explanation:

The macrocyclic effect in coordination chemistry refers to the enhanced stability of metal complexes that have a macrocyclic ligand. A macrocyclic ligand is a ligand that forms a ring around the metal ion, forming a macrocyclic complex. This structure increases the enthalpy of the complex, making it more thermodynamically stable. In other words, the macrocyclic effect increases the stability of a metal complex by making the ligand more difficult to remove. This effect is especially important in biological systems, where macrocyclic ligands play a key role in many enzymatic reactions.

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Object A has a molar heat of 31.2 J/mole∙°C and object B molar heat is 11.2 J/mole∙°C. Which object will heat up faster if they have the same mass and equal amount of heat is applied? Explain why.
Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. Express the cell potential to two decimal places and include the appropriate units.

A solution of HNO3HNO3 is standardized by reaction with pure sodium carbonate. 2H++Na2CO3⟶2Na++H2O+CO2 2H++Na2CO3⟶2Na++H2O+CO2 A volume of 27.71±0.05 mL27.71±0.05 mL of HNO3HNO3 solution was required for complete reaction with 0.9585±0.0007 g0.9585±0.0007 g of Na2CO3Na2CO3 , (FM 105.988±0.001 g/mol105.988±0.001 g/mol ). Find the molarity of the HNO3HNO3 solution and its absolute uncertainty.

Answers

Answer:

(0,653±0,002) M of HNO₃

Explanation:

The reaction of standarization of HNO₃ with Na₂CO₃ is:

2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻

To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (27,71±0,05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus:

0,9585 g of Na₂CO₃ × ( 1 mole / 105,988 g) =

9,043×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,809×10⁻² moles of HNO₃

Molarity is moles divide liters, thus, molarity of HNO₃ is:

1,809×10⁻² moles / 0,02771 L = 0,6527 M of HNO₃

The absolute uncertainty of multiplication is the sum of relative uncertainty, thus:

ΔM = 0,6527M× (0,0007/0,9585 + 0,001/105,988 + 0,05/27,71) =

0,6527 M× 2,54×10⁻³ = 1,7×10⁻³ M

Thus, molarity of HNO₃ solution and its absolute uncertainty is:

(0,653±0,002) M of HNO₃

I hope it helps!

The standard entropy of Pb(s) at 298.15 K is 64.80 J K–1 mol–1. Assume that the heat capacity of Pb(s) is given by: CP,m(Pb, s) J K−1mol−1 = 22.13 + 0.01172 T K + 1.00 x 10−5 T 2 K2 The melting point is 327.4 ℃ and the heat of fusion is 4770 J mol-1. Assume that the heat capacity of Pb(l) is given by: CP,m(Pb, l) J K−1mol−1 = 32.51 − 0.00301 T K Calculate the standard entropy of Pb(l) at 500 ℃

Answers

Answer:

$s_(Pb(l),500\°C)=100.83(J)/(mol*K)$

Explanation:

Hello,

In this case, for the calculation of the standard entropy of liquid lead at 500 °C (773.15 K), starting by solid lead 298.15 K we need to consider three processes:

1. Heating of solid lead at 298.15 K to 600.55 K (melting point).

2. Melting of solid lead to liquid lead.

3. Heating of liquid lead at 600.55 K (melting point) to 773.15 K.

Which can be written in terms of entropy by:

$s_(Pb(l),500\°C)=s_(Pb(s),298.15K)+s_1+s_2+s_3$

Whereas each entropy is computed as follows:

$s_1=\int\limits^(600.55K)_(298.15K) {(22.13 + 0.01172 T + 1.00 x 10^(-5) T^2)/(T) } \, dT =20.4(J)/(mol*K)\n\n\ns_2=(4770(J)/(mol) )/(600.55K)= 7.94(J)/(mol*K)\n\n\ns_3=\int\limits^(773.15K)_(600.55K) {(32.51-0.00301T)/(T) } \, dT=7.69(J)/(mol*K)$

Therefore, the standard entropy of liquid lead at 500 °C turns out:

$s_(Pb(l),500\°C)=64.80+20.4+7.94+7.69\n\ns_(Pb(l),500\°C)=100.83(J)/(mol*K)$

Best regards.

You just measured a metal cylinder and obtained the following information: mass - 3.543 g diameter -0.53 cm height = 4.40 cm. 265 • Determine the volume (V). (V=nrºh, where r = radius, h = height, T - 3.14) V= (3.14 36205² 4.402.17cm? • Determine density of the cylinder D. 3. SMS-365923 197

Answers

The density of the cylinder would be 3.652 gram/ cm³

What is density?

It can be defined as the mass of any object or body per unit volume of the particular object or body. Generally, it is expressed as in gram per cm³ or kilogram per meter³.

As given in the problem, you just measured a metal cylinder and obtained the following information: mass - 3.543 grams, diameter 0.53 cm, height = 4.40 cm , and we have to calculate the density of the cylinder,

mass of the cylinder = 3.543 grams

the volume of the cylinder = πr²h

= 3.14 ×.265²×4.4

=0.97 cm³

By using the above formula for density

ρ = mass of the cylinder/volume of the cylinder

= 3.543 grams/0.97 cm³

=3.652 grams/ cm³

Thus,the density of the cylinder would be 3.652 grams/ cm³.

To learn more about density from here, refer to the link;

brainly.com/question/15164682

#SPJ2

Answer:

V cylinder = $\pi r^2 h = \pi(0.53 cm/2)^24.4cm= 0.97 cm^3$

note that $r=diameter/2$

density = $m/V= 3.543/0.97= 3.65 g/cm^3$

What is colloidal solutions

Answers

Explanation:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid. ... Colloidal systems can occur in any of the three key states of matter gas, liquid or solid. However, a colloidal solution usually refers to a liquid concoction.

Answer:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid.

Consider the following reaction at equilibrium: NO2(g) + CO(g) = NO(g) + CO2(g) Suppose the volume of the system is decreased at constant temperature, what change will this cause in the system? A shift to produce more NO A shift to produce more CO A shift to produce more NO2 No shift will occur

Answers

Answer: Option (d) is the correct answer.

Explanation:

According to Le Chaltelier's principle, when there occurs any change in an equilibrium reaction then the equilibrium will shift in a direction that will oppose the change.

This means that when pressure is applied on reactant side with more number of moles then the equilibrium will shift on product side that has less number of moles.

For example, $NO_(2)(g) + CO(g) \rightleftharpoons NO(g) + CO_(2)(g)$

Since here, there are same number of moles on both reactant and product side. So, when volume is decreased at a constant temperature in this system then there will occur no change in the equilibrium state.

Thus, we can conclude that in the given when volume of the system is decreased at constant temperature, then no shift will occur.

lucose, a major energy-yielding nutrient, is present in bacterial cells at a concentration of approximately 0.200 mM. i) What is the concentration of glucose in the E. coli cell in mg/mL?

Answers

Answer:

The concentration is 0.036 mg/mL

Explanation:

Concentration = 0.2 mM = 0.2/1000 = 2×10^-4 M = 2×10^-4 mol/L × 180,000 mg/1 mol × 1 L/1000 mL = 0.036 mg/mL

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