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faltersainse [42]

2 years ago

11

i really need the answer please i will give 30 points to the person that answers it i swear to god i put my hand on the bible Pa

tricia has gone fishing with her son. Which marine zone would she be able to fish in using a hook?

2 answers:

Alecsey [184]2 years ago

8 0

The photic zone and the mesopelagic

MrRa [10]2 years ago

3 0

The Marine zones I believe.

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

cupoosta [38]

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.

8 0

3 years ago

Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina

yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

8 0

2 years ago

HELP PLZ !!! 15 points

Ulleksa [173]

Hi I’m here, what can I do to help

3 0

3 years ago

Read 2 more answers

(2pts) During the Purification of Lactate Dehydrogenase (LDH) experiment, you will need 50ml of buffer A150. Buffer A150 is 30mM

torisob [31]

Answer:

The answer is " $20 \ mL$ "

Explanation:

Given:

Molarity= number of moles

because it is 1 Liter

$\to \frac{0.03\ moles}{1.5 moles}=0.02\ L= 20 \ mL \ of\ Tris\\\\$

therefore,

it takes 20 mL of Tris.

$\to \frac{0.150 \ moles}{5\ moles} =0.03\ L\\\\$

$= 30 \ mL \ of\ Nacl$

So, take $20 \ mL\ of\ NaCl.$

6 0

2 years ago

Using the standard reduction potentials, Pb 2+(aq) + 2e– => Pb(s), E° = –0.13 V Fe 2+(aq) + 2e– => Fe(s), E° = –0.44 V Zn

mojhsa [17]

Answer:

Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$

Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$

Explanation:

Standard reduction potential denotes ability to consume electrons from another species.

Hence, higher the standard reduction potential, higher will be the ability to oxidize another species.

Metal with $E_{red}^{0}$ value lower than 1.51 V will donate electron to $Mn^{3+}$ and thus reduces $Mn^{3+}$ to $Mn^{2+}$ .

So, Pb(s), Fe(s) and Zn(s) will reduce $Mn^{3+}$ to $Mn^{2+}$ .

Metal with $E_{red}^{0}$ value lower than -0.40 V will donate electron to $Cr^{3+}$ and thus reduces $Cr^{3+}$ to $Cr$ .

So, Fe(s) and Zn(s) will reduce $Cr^{3+}$ to $Cr$ .

6 0

2 years ago