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BigorU [14]
3 years ago
5

Why the measured pressure of a gas under conditions that are very close to those that would result in condensation will be lower

than what the ideal gas law would predict?
Chemistry
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer:

Inter-molecular forces and molecular volumes are the chief reasons for lower measured pressure

Explanation:

The kinetic theory assumes that gas particles occupy a negligible fraction of the total volume of the gas. It also assumes that the force of attraction between gas molecules is zero.

However, during high pressure, the volume of the gas particles are not negligible compare to the total gas volume and as such the volume of a real gas under such condition is higher than the Ideal gas. Vander-waal attempted to modify the ideal gas equation by subtracting the excess volume from the ideal equation. The increased volume is the reason the measured pressure of a real gas is  less than an ideal gas

On the other hand,  close to condensation, the other  assumption of negligible forces of attraction becomes invalid. As inter-molecular distances decrease, inter-molecular forces increase reducing the bombardment of the wall of the container due to restricted particle movement and lower measured gas pressure.

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Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h
Genrish500 [490]

Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

k=\frac{0.693}{t_1_/_2}

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

k=\frac{0.693}{1.5 hrs}

The expression for the rate constant is :

k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

5 0
3 years ago
A substance that increases the rate of a chemical reaction is called a(n) _____. answer
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Catalyst increases the rate of a chemical reaction
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3 years ago
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CO2(g) + H2O(I) - -> C6H12O6(s) + O2(g)
Lana71 [14]

Answer:

For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced, but PhC2H5 + O2 = PhOH + CO2 + H2O will; Compound states [like (s) (aq) or (g)] are not required. If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.

Explanation:

4 0
3 years ago
You carefully weigh out 20.00 g of CaCO3 powder and add it to 81.00 g of HCl solution. You notice bubbles as a reaction takes pl
Zepler [3.9K]

Answer:

The relevant equation is:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Explanation:

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The formed CO₂ is the reason why you noticed bubbles as the reaction took place

3 0
3 years ago
What is the change in internal energy ( ΔU ) of the system if q = –8 kJ and w = –1 kJ for a certain process?
Radda [10]

Answer:

Change in internal energy (ΔU) = -9 KJ

Explanation:

Given:

q = –8 kJ [Heat removed]

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Find:

Change in internal energy (ΔU)

Computation:

Change in internal energy (ΔU) = q + w

Change in internal energy (ΔU) = -8 KJ + (-1 KJ)

Change in internal energy (ΔU) = -8 KJ - 1 KJ

Change in internal energy (ΔU) = -9 KJ

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3 years ago
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