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Lostsunrise [7]
3 years ago
14

At 25 degrees, the rate constant for the first order decomposition of a pesticide solution is 6.40 x 10^-2 min^-2. If the starti

ng concentration of pesticide is 0.0314 M, what concentration will remain after 62.0 min at 25 degree Celsius?
The correct answer is 2.11 x 10^2 M, please show the steps.
Chemistry
1 answer:
EastWind [94]3 years ago
7 0
<span>If the initial amount is 1, the amount remaining at time t is exp(-rt), where r is the rate constant (.0064), and t is the time (62). The value of the exponential is 0.6725, and doing the obvious multiplication gives .0211 mole.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
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I’m going to guess but B?
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3 years ago
Before tackling this problem, be sure you know how to find the antilog of a number using a scientific calculator.
dybincka [34]
<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>
  1. pH:- 5.4
  2. pH = - log[H+]

<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

Antilog (-6)  = {10}^{ - 6} ,  \\ Antilog (0.6)  = 4

put the value in equation

{10}^{ - 6}   \times 4 = [H+] \\ 4 \times   {10}^{ - 6}  = [H+]

7 0
2 years ago
Read 2 more answers
1. If I have 45 L of He in a balloon at 25 degrees celsius and increase the temperature of the
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Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).

We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.

Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.  

(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).

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3 years ago
Round each number to four significant figures
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Answer:

a.

84,791 » 8479

b.

256.75 » 256.8

c.

431,801 » 4318

d.

0.00078100 » 0.0007810

Explanation:

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