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In the reaction 5 space B r to the power of minus space (a q )space plus space B r O subscript 3 to the power of minus space (a q )space plus space 6 space H to the power of plus space (a q )space rightwards arrow space 3 space B r subscript 2 space (a q )space plus space 3 space H subscript 2 O space (l )the rate of disappearance of Br- at some moment in time was determined to be 3.5 x 10-4 M/s. What is the rate of appearance of Br2 at that same moment

Answers

Answer 1

Answer:

Answer:

$r_(Br_2)=2.1x10^(-4)M/s$

Explanation:

Hello,

In this case, for the reaction:

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

Thus, via the rate proportions between Br⁻ and Br₂ for which the stoichiometric coefficients are 5 and 3 respectively, we can write:

$(r_(Br^-))/(-5) =(r_(Br_2))/(3)$

Hence, the rate of appearance of Br₂ turns out:

$r_(Br_2)=(3r_(Br^-))/(-5)=(3*-3.5x10^(-4)M/s)/(-5)\n \nr_(Br_2)=2.1x10^(-4)M/s$

Take into account that the rate of disappearance is negative for reactants.

Best regards.

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A piston confines 0.200 mol Ne(g) in 1.20 at 25 degree C. Two experiments are performed. (a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm. (b) The gas is allowed to expand reversibly and isothermally to the same final volume. Please calculate the work done by the gas system in these two processes, respectively. Which process does more work? (revised from 6/e exercise 8.11) Please show calculation details.

Answers

Answer:

The second experiment (reversible path) does more work

Explanation:

Step 1:

A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C

(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm

Irreversible path: w =-Pex*ΔV

⇒ with Pex = 1.00 atm

⇒ with ΔV = 1.20 L

W = -(1.00 atm) * 1.20 L

W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J

(b) The gas is allowed to expand reversibly and isothermally to the same final volume.

W = -nRTln(Vfinal/Vinitial)

⇒ with n = the number of moles = 0.200

⇒ with R = gas constant = 8.3145 J/K*mol

⇒ with T = 298 Kelvin

⇒ with Vfinal/Vinitial = 2.40/1.20 = 2

W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)

W = -343.5 J

The second experiment (reversible path) does more work

Select the correct answer.In which research method does a researcher change the value of one variable to determine how the change affects another variable?
O A experiment
OB. observational research
Oc.
survey
OD
case study

Answers

Answer:

experiment is the answer

The following reaction was monitored as a function of time: A→B+C A plot of ln[A] versus time yields a straight line with slope −4.3×10−3 /s. If the initial concentration of A is 0.260 M, what is the concentration after 225 s?

Answers

The concentration after 225 s is 0.099 M.

As we know that, the graph of ln [A] versus time yields a straight line with slope 'k'.

So, Slope = k = $4.3*10^(-3)/s$

Rate law for first order kinetics:

$t=(2.303)/(k) log (a)/(a-x)$

where,

k = rate constant = $4.3*10^(-3)/s$

t = time passed by the sample = 225 s

a = initial amount of the reactant = 0.260 M

a - x = amount left after decay process = ?

On substituting the values:

$t=(2.303)/(k) log (a)/(a-x)\n\nt=(2.303)/(4.3*10^(-3)) log (0.260)/(a-x)\n\na-x=0.099M$

Therefore, the concentration after 225 s is 0.099 M.

Find more information about Rate law here:

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Answer : The concentration after 225 s is, 0.099 M

Explanation :

As we know that, the graph of ln [A] versus time yields a straight line with slope 'k'.

So, Slope = k = $4.3* 10^(-3)s^(-1)$

Expression for rate law for first order kinetics is given by:

$t=(2.303)/(k)\log(a)/(a-x)$

where,

k = rate constant = $4.3* 10^(-3)s^(-1)$

t = time passed by the sample = 225 s

a = initial amount of the reactant = 0.260 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$225=(2.303)/(4.3* 10^(-3))\log(0.260)/(a-x)$

$a-x=0.099M$

Therefore, the concentration after 225 s is, 0.099 M

A range of organic molecules can undergo combustion. If pyridine

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Answer:

A range of organic molecules can undergo combustion. Pyridine (C5H5 combustion in the unbalanced reaction shown below wtar o 4 CsH5N + O2 +H2O + CO2 + NO a) Write the balanced equation. (2 points) # 41 CH N +170 70 the flow, t- b) Find the percent yield for the reaction if 10.0 g of pyridine dioxide. (2 points)

Explanation:

hope this helps!

Which issue is a limitation of using synthetic polymers

Answers

The main issues of using synthetic polymers include toxicity poor biocompatibility etc. Synthetic polymers stay non-degradable for ling time and make the surface polluted.

What are synthetic polymers?

Natural polymers are naturally made substances such as cellulose, starch, glycogen etc. Polymers made by man are called synthetic polymers. Synthetic polymers are diverse and are made through several polymerization techniques.

PVC, polyethylene, polyesters Teflon etc. are very common polymers in daily life. A major class of synthetic polymers include plastics which are major pollutants nowadays.

Most of the synthetic polymers are non-biodegradable and will cause landfill issues. Some them are toxic in nature and might cause several health issues. Blending them with biodegradable polymers is a solution for this.

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Answer: As trash, Synthetic Polymers are not biodegradable. Landfills can easily fill up with synthetic polymers. Plastics can be made into different products. Recycling synthetic polymers is costly.

Explanation: Hope this helps in any way possible!

Which energy conversion occurs in an operating voltaic cell? A) electrical energy to nuclear energy B) chemical energy to electrical energy C) chemical energy to nuclear energy D) electrical energy to chemical energy

Answers

The correct option is B.

Which process occurs in a voltaic cell?

A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy. The important parts of a voltaic cell: The anode is an electrode where oxidation occurs. The cathode is an electrode where reduction occurs.

What kind of energy is converted in a galvanic cell?

A galvanic cell is an electrochemical cell that converts the free energy of a chemical process into electrical energy. A photogalvanic cell is one that generates species photochemically which react resulting in an electrical current through an external circuit.

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Answer:

B) chemical energy to electrical energy

Explanation:

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