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yan [13]
3 years ago
15

33.23 grams of a thin sheet of iron is completely oxidized in 7 days. How would you express the rate of conversion of the silver

?
.0440 moles/day
.183 moles/hour
.0440 moles/week
.308 moles/day
Chemistry
1 answer:
tresset_1 [31]3 years ago
7 0

Answer:

A. 0.0440 moles/day

Explanation:

First, let's figure out how many moles 33.23 grams of silver is. We do this by dividing the number of grams by the molar mass of silver, which is 107.87 g/mol:

33.23 g Ag ÷ 107.87 g/mol = 0.3081 mol Ag

Now, let's divide this by 7 to get the rate per day:

0.3081 mol Ag ÷ 7 days = 0.0440 mol/day

Thus, the answer is A.

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20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator
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Answer:

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

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Explanation:

The equation for the reaction is :

C_2H_5NH_2_(_a_q_)     +     H^+_(_a_q_)   ---      C_2H_5NH_{3(aq)}^+

concentration of C_2H_5NH_{2(aq) = 10%

10 g of C_2H_5NH_{2(aq) in 100 ml solution

molar mass = 45.08 g/mol

number of moles = 10 / 45.08

= 0.222 mol

Molarity of C_2H_5NH_2(aq) = 0.222 \times \frac{1000}{100}mL

= 2.22 M

number of moles of C_2H_5NH_{2(aq) in 20 mL can be determined as:

= 20 mL \times  2.22 M= 44*10^{-3} mole

Concentration of C_2H_5NH_2(aq) = \frac{44*10^{-3}*1000}{20}

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Similarly, The pKa Value of C_2H_5NH_{2(aq) is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

the pH value at equivalence point is,

pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]

pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]

pH = 5.21

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