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maria [59]
3 years ago
12

What is the rate constant of a first-order reaction that takes 5.50 minutes for the reactant concentration to drop to half of it

s initial value?
Chemistry
1 answer:
bulgar [2K]3 years ago
8 0
For a first order reaction, the half life is inversely proportional to the rate constant. 
The formula is
half life = ln(2)/k = 0.693/k
where k is the rate constant

t = 5.50 minutes

k = ln(2)/5.50 = 0.126 min^-1

Your rate constant is 0.126 min^-1.

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Fe(CN)63- + Re → Fe(CN)64- + ReO4--. What is reduced?
givi [52]

Answer:

1. Fe is reduced

2. Mn is Oxidized

3. N is oxidized

Explanation:

<em>Check the image below:</em>

Reducing agent is an element or compound that loses an electron to an electron recipient in a redox chemical reaction. oxidizing agent is a substance that has the ability to oxidize other substances — in other words to accept their electrons.

6 0
3 years ago
The active form of vitamin D is called ________.
alexandr1967 [171]
<span>Calcitriol </span>is the active form of vitamin D. (D)
7 0
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Read 2 more answers
What is irrigation?Describe two methods of irrigation which conserve water​
MA_775_DIABLO [31]

The process of watering the crops is called irrigation.

Any two methods of irrigation are:

<h3>(i) Sprinkler system:</h3>

This system is used on the uneven land where less water is available. The perpendicular pipes, having rotating nozzles on top, are joined to the main pipeline at regular intervals. Water is allowed to flow through main pipe under pressure, which escapes from the rotating nozzles. In this way water gets sprinkled on the crop.

<h3> (ii) Drip irrigation:</h3>

This system is used to save water as it allows the water to flow drop by drop at the roots of the plants. It is the best technique for watering fruit plants, gardens and trees. Water is not wasted at all.

5 0
3 years ago
1. If a car is traveling 90 mi/hr, how many feet will the car travel in 1 sec? In 5 sec?
Fudgin [204]

Answer:

The car travel 660 feet

Explanation:

First convert the Speed into <u>Feet/sec</u>

Speed of the car = 90 mi/hr

1 mi = 5280 ft

In 90 miles = 90 x 5280 = 475,200  feet

1 hr = 60 min

1 min = 60 sec

So ,

1 hr = 60 x 60 sec = 3600 sec

The speed is defined as the distance traveled by the object in unit time. The formula of speed is :

Speed =\frac{distance}{time}

Speed =\frac{90mi}{1hr}... given

Speed =\frac{475200 ft}{3600sec}

Speed = 132 ft/sec

Now,  time = 5 sec

Speed =\frac{distance}{time}

distance =speed\times time

distance = 132ft/sec\times 5sec

Distance = 660 feet

3 0
3 years ago
Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M acetic acid (Ka = 1.75x10-5) with 0.150
mylen [45]

Answer:

a) pH = 2.793

b) pH = 4.280

c) pH = 4.933

d) pH = 8.816

e) pH = 8.861

f) pH = 8.891

Explanation:

a) VNaOH = 0 mL

∴ CH3COOH ↔ CHECOO- + H3O+

⇒ Ka = 1.75 E-5 = [ H3O+ ] * [ CH3COO-] / [ CH3COOH ]

mass balance:

⇒ <em>C</em> CH3COOH = [ CH3COO- ] + [ CH3COOH ] = 0.150 M

charge balance:

⇒ [ H3O+ ] = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ]² / ( 0.150 M - [ H3O+ ] )

⇒ [ H3O+ ]² + 1.75 E-5 [ H3O+ ] - 2.625 E-6 = 0

⇒ [ H3O+ ] = 1.61146 E-3 M

⇒ pH = - Log [ H3O+ ] = 2.793

b) after  5.0 mL NaOH:

∴ CH3COOH + NaOH ↔ CH3COONa + H2O

⇒ <em>C</em> NaOH = (5 E-3 L * 0.150 mol/L) / (0.025+0.01 ) = 0.02143 M

⇒ <em>C</em> CH3COOH = ((0.025*0.150) - (0.01*0.150)) / (0.025 + 0.01) = 0.0643 M

mass balance:

⇒ 0.02143 + 0.0643 = [ CH3COOH ] + [ CH3COO- ] = 0.086 M

charge balance:

⇒ [ H3O+ ] + [Na ] = [ CH3COO- ]

⇒ [ H3O+ ] + 0.02143 = [ CH3COO- ]

⇒ Ka = [ H3O+ ] * ( [ H3O+ ] + 0.150 ) / (0.086 - 0.02143 - [ H3O+ ]) = 1.75 E-5

⇒ [ H3O+ ]² + 0.02143 [ H3O+ ] = 1.13 E-6 - 1.75 E-5 [ H3O+ ]

⇒ [ H3O+ ]² + 0.02144 [ H3O+ ] - 1.13 E-6 = 0

⇒ [ H3O+ ] = 5.26 E-5 M

⇒ pH = 4.28

c) after 15 mL NaOH:

⇒ <em>C</em> CH3COOH = 0.0375 M

⇒ <em>C</em> NaOH = 0.05625 M

mass balance:

⇒ 0.09375 M = [ CH3COO- ] +[ CH3COOH ]

charge balance:

⇒ [ H3O+ ] + 0.05625 = [ CH3COO- ]

⇒ Ka = 1.75 E-5 = [ H3O+ ] * ([ H3O+ ] + 0.05625) / (0.09375 - 0.05625 - [H3O+])

⇒ [H3O+]² + 0.05625[H3O+] = 6.5625 E-7 - 1.75 E-5 [H3O+]

⇒ [ H3O+]² + 0.05626[H3O+] - 6.5625 E-7 = 0

⇒ [ H3O+ ] = 1.1662 E-5 M

⇒ pH = 4.933

d) after 25 mL NaOH:

⇒ <em>C </em>NaOH = 0.075 M

⇒ <em>C</em> CH3COOH = 0 M....equiv. point

⇒Kh = Kw/Ka = 1 E-14 / 1.75 E-8 = 5.7143 E-10 = [ OH-]² / ( 0.075 - [OH-])

⇒ [OH-]² + 5.7143 E-10[OH-] - 4.286 E-11 = 0

⇒ [ OH- ] = 6.5463 E-6 M

⇒ pOH = 5.184

⇒ pH = 8.816

e) after 40 mL NaOH:

⇒ <em>C </em>NaOH = 0.0923 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 5.275 E-11 = 0

⇒ [OH-] = 7.2624 E-6 M

⇒ pOH = 5.139

⇒ pH = 8.861

f) after 60 mL NaOH:

⇒ <em>C </em>NaOH = 0.106 M

⇒ [OH-]² + 5.7143 E-10 [OH-] - 6.05 E-11 = 0

⇒ [OH-] = 7.7782 E-6 M

⇒ pOH = 5.11

⇒ pH = 8.891

5 0
3 years ago
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