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WITCHER [35]
3 years ago
13

A student conducting the iodine clock experiment accidentally makes an s2o32- stock solution that is too concentrated. how will

this affect the rate measurement? reaction 1: 3i- (aq) + s2o82- (aq) -> i3- (aq) + 2so42- (aq) slow reaction 2: i3- (aq) + 2s2o32- (aq) -> 3i- (aq) + s4o62- (aq) fast reaction 3: i3- (aq) + starch (aq) -> 3i- ---- starch (bluish black) fast
Chemistry
1 answer:
cupoosta [38]3 years ago
8 0
Adding (S2O3)2- would affect the reaction mechanism that involves this ion. From the reaction mechanism given above, the equilibrium of step 2 would be affected. Adding the stock solution of (S2O3)2- would shift the equilibrium to the right thus making more products of the said mechanism. Also, the reaction rate of this step would occur faster than the original rate. This is based on Le Chatelier's Prinicple which states that a corresponding change would happen to the equilibrium of a reaction when pressure, concentration of the substances or temperature is changed. So, that after the addition, a color change would appear immediately because I3- would be removed slowly from solution, and would therefore be able to react with starch. 

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Cobalt-60 is a strong gamma emitter that has a half- life of 5.26 yr. The co balt-60 in a radiotherapy unit must be replaced whe
Alenkasestr [34]

<u>Answer:</u> The sample of Cobalt-60 isotope must be replaced in January 2027

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

where,

t_{1/2} = half life of the reaction = 5.26 years

Putting values in above equation, we get:

k=\frac{0.693}{5.26yrs}=0.132yrs^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 0.132yr^{-1}

t = time taken for decay process = ? yr

[A_o] = initial amount of the sample = 100 grams

[A] = amount left after decay process =  (100 - 75) = 25 grams

Putting values in above equation, we get:

0.132=\frac{2.303}{t}\log\frac{100}{25}\\\\t=10.5yrs

The original sample was purchased in June 2016

As, June is the 6th month of the year, which means the time period will be 2016+\frac{6}{12}=2016.5

Adding the time in the original time period, we get:

2016.5+10.5=2027

Hence, the sample of Cobalt-60 isotope must be replaced in January 2027

3 0
3 years ago
What does the chemical formula CaCI2 show about the compound or represents?
atroni [7]

Answer:

Two elements because Can is one element and Cl is another. The two is a coeficient that states how many of that element there is.

I hope this helps you!!!

8 0
3 years ago
Why was Newland’s periodic table rejected?
Rina8888 [55]
He put iron with sulfur and oxygen and it had a few more errors and iron is a metal the other two are nonmetals

5 0
3 years ago
Kc for the reaction N2O4 &lt;=&gt; 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w
Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619=  [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373  

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10  = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0
3 years ago
WILL MARK BRAINLIEST
den301095 [7]
The answer is C. Life
Hope this helps! :)
7 0
3 years ago
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