A sample of phosgene gas at an initial concentration of 0.500 m is heated at 527 °c in a reaction vessel. at equilibrium, the concentration of co (g) was found to be 0.046 m. calculate the equilibrium constant for the reaction at 527 °c.

Answers

Answer 1

Answer: Answer : The equilibrium constant will be - 0.454

Explanation : The reaction is given below;

$COCl_(2) \ \textless \ ----\ \textgreater \ CO + Cl_(2)$

We need to find Equilibrium constant - $K_(c)$ ;

$K_(c)$ = [CO] [ $Cl_(2)$ ] / $[COCl_(2) ]$

So, $K_(c)$ = [0.046] X [0.046] / [0.5 - 0.046]

Therefore, $K_(c)$ = 0.454

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Under what set of conditions does HCl(g) deviate the most from ideal behavior?a) high temperature and low pressure.b) low temperature and high pressure.c) high temperature and high pressure.d) low temperature and low pressure. e) standard temperatore and pressure.

Answers

Answer:

b) low temperature and high pressure

Explanation:

Deviation of gases from their ideal behavior could be as a result of two things which include a very small volume for the gases and the collisions not being elastic enough.

Small volume will decrease the frequency of elastic collisions a gas will experience as a result of the tiny space.

Low temperature will decrease the amount of elastic collisions and energy of the gas and high pressure will decrease the volume which is why there will be a deviation in the ideal behavior of Hcl(g)

When an ionic compound such as sodium chloride (NaCl) is placed in water, the component atoms of the NaCl crystal dissociate into individual sodium ions (Na⁺) and chloride ions (Cl-). In contrast, the atoms of covalently bonded molecules (e.g. glucose, sucrose, glycerol) do not generally dissociate when placed in aqueous solution. Which of the following solutions would be expected to contain the greatest number of solute particles (molecules or ions)?A) 1 litre of 0.5 M NaClB) 1 litre of 1.0 M NaClC) 1 litre of 1.0 M glucoseD) 1 litre of 1.0 M NaCl and 1 litre of 1.0 M glucose will contain equal numbers of solute particles.

Answers

Answer:

1 litre of 1.0 M NaCl

Explanation:

When an ionic compound dissolves in water, it dissociates into ions. Consider the dissolution of sodium chloride in water;

NaCl(s) ------> Na^+(aq) + Cl^-(aq)

Hence, two solute particles are obtained from each formula unit of NaCl, a greater concentration of NaCl will contain a greater number of sodium an chloride ion particles.

Glucose is a molecular substance and does not dissociate in solution hence it yields a lesser number of particles in solution even at the same concentration as NaCl

Final answer:

The solution with the greatest number of solute particles is 1 litre of 1.0 M NaCl, as ionic compounds dissociate into individual ions, thus providing more particles per litre.

Explanation:

Given the details of the question, the solution that would be expected to contain the greatest number of solute particles would be 1 litre of 1.0 M NaCl. This is because when ionic compounds like sodium chloride are placed in water, they dissociate into individual ions. In the case of NaCl, it splits into two ions, sodium (Na+) and chloride (Cl-). Thus, a 1.0 M solution of NaCl would actually contain 2.0 moles of particles per litre because each formula unit of NaCl gives two particles. Covalently bonded molecules like glucose do not dissociate in solution, therefore, a 1.0 M glucose solution would have 1.0 mole of particles per litre.

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The following data is given to you about a reaction you are studying: Overall reaction: 2A  D Proposed mechanism: Step 1 A + B  C (slow) Step 2 C + A  D + B (fast) [A]o = 0.500 M [B]o = 0.0500 M [C]o = 0.500 M [D]o = 1.50 M This reaction was run at a series of temperatures and it was found that a plot of ln(k) vs 1/T (K) gives a straight line with a slope of -982.7 and a Y intercept of -0.0726. What is the initial rate of the reaction at 298K?

Answers

Answer : The initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

Explanation :

The Arrhenius equation is written as:

$K=A* e^{(-Ea)/(RT)}$

Taking logarithm on both the sides, we get:

$\ln k=-(Ea)/(RT)+\ln A$ ............(1)

where,

k = rate constant

Ea = activation energy

T = temperature

R = gas constant = 8.314 J/K.mole

A = pre-exponential factor

The equation (1) is of the form of, y = mx + c i.e, the equation of a straight line.

Thus, if we plot a graph of $\ln k$ vs $(1)/(T)$ then the graph shows a straight line with negative slope. That means,

Slope of the line = $-(Ea)/(R)$

And,

Intercept = $\ln A$

As we are given that:

Slope of the line = -982.7 = $-(Ea)/(R)$

Intercept = -0.0726 = $\ln A$

Now we have to calculate the value of rate constant by putting the value of slope, intercept and temperature (298K) in equation 1, we get:

$\ln k=-(982.7)/(298)+(-0.0726)$

$\ln k=-3.37$

$k=0.0344s^(-1)$

The value of rate constant is, $0.0344s^(-1)$

Now we have to calculate the initial rate of the reaction at 298 K.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$A+B\rightarrow C$

The expression of rate law for this reaction will be,

$Rate=k[A][B]$

As we are given that:

[A] = 0.500 M

[B] = 0.0500 M

k = $0.0344s^(-1)$

Now put all the given values in the rate law expression, we get:

$Rate=(0.0344)* (0.500)* (0.0500)$

$Rate= 8.6* 10^(-4)M/s$

Therefore, the initial rate of the reaction at 298 K is, $8.6* 10^(-4)M/s$

To what extend must a soln conc 40mg 4g NO3 pa cm3 be dilute to yield one of conc 16mg 4g NO3 pa cm3

Answers

Each ml should be dilute to 2.5 ml

To each ml of solution 1.5 ml of water should be added

A piece of charcoal used for cooking is found at the remains of an ancient campsite. a 0.94 kg sample of carbon from the wood has an activity of 1580 decays per minute. find the age of the charcoal. living material has an activity of 15 decays/minute per gram of carbon present and the half-life of 14c is 5730 y. answer

Answers

Mass of sample of charcoal = 0.94 kg = 0.00094

∴, activity = decay rate / mass = 1580/0.00094
= 1.681 X 10^6 decays per min per gram

Using the half-life formula, we have:
activity of sample / activity of modern carbon = (1/2)^(age / half-life)
∴, Age = half-life x log (base 2) (modern activity / coal activity)
= 5730 x log(base 2)(1.681X10^6/ 15)
= 96115 years.

Answer: Age of the charcoal = 96115 years

Final answer:

Using the radiocarbon dating technique and applying the decay formula, it is calculated that the age of the charcoal from the an ancient campsite is approximately 9,500 years.

Explanation:

The age of the charcoal can be found using the technique of radiocarbon dating, which capitalizes on the process of radioactive decay. The isotope carbon-14 (¹4C) is used in this method as it has a known half-life of 5730 years. The number of decays per minute per gram of carbon in a live organism is known as its activity.

Initially, the activity was given as 15 decays per minute per gram. The present activity of the carbon in the charcoal is provided at 1580 decays per minute for a 0.94 kg or 940 gram sample. Thus, the current activity per gram is 1580/940 equals approximately 1.68 decays per minute per gram.

Given that the half-life of ¹4C is 5730 years, we can apply the formula for calculating the time passed using the rate of decay, which is given as T = (t1/2 / ln(2)) * ln(N0/N), where 'ln' is the natural logarithm, 'N0' is the initial quantity (15 decays/minute per gram), 'N' is the remaining quantity (1.68 decays/minute per gram).

Plugging in the given values, we get T = (5730 / ln(2)) * ln(15/1.68), which gives us approximately 9,500 years. Therefore, the age of the charcoal is around 9,500 years.

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A potassium ion (K+) would most likely bond with _____.

Mg+
O
Cl-
Na+

Answers

Answer:

Chlorine

Explanation:

Answer:

That would be the negative ion: Cl-.

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