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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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What type of bond will form between two chlorine atoms?
Trava [24]
Since both atoms are the same and are both nonmetals, they would form a Nonpolar covalent bond. This bond occurs when usually atoms of the same element or atoms of propriety electronegativity differences are sharing electrons to form bonds. There is an equal sharing of valence electrons in this chemical bond.
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Why are all atoms electrically neutral?
Anna11 [10]

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7 0
3 years ago
35.0 grams of nitrogen gas reacts with 60.0 grams of hydrogen gas: N2 + 3H2--&gt; 2NH3
Misha Larkins [42]

Explanation:

Moles of N2 = 35.0g / (28g/mol) = 1.25mol

Moles of H2 = 60.0g / (2g/mol) = 30.0mol

Since 1.25mol * 3 < 30.0mol, nitrogen is limiting.

Moles of NH3 = 1.25mol * 2 = 2.50mol.

Mass of NH3 = 2.50mol * (17g/mol) = 42.5g.

30.0mol - 1.25mol * 3 = 26.25mol.

Excess mass of H2

= 26.25mol * (2g/mol) = 52.5g.

6 0
3 years ago
Balance the Chemical Equations<br><br> MgF2 + Li2CO3---&gt; MgCO3 + LiF <br><br> Please Help
kirill115 [55]

Answer:

MgF2 + Li2CO3 ---> MgCO3 + 2LiF

Explanation:

4 0
3 years ago
Calculate the pH of the solution that results when 20 mL of 0.2M HCOOH is mixed with 25mL of 0.2M NaOH solution.(post-equivalenc
Paul [167]

The pH of the solution : 12

<h3>Further explanation</h3>

Reaction

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

mol HCOOH =

\tt 20~ml\times 0.2~M=4~mlmol

mol NaOH =

\tt 25~ml\times 0.2~M=5~mlmol

Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]

ICE method :

HCOOH    +     NaOH   ⇒     HCOONa   +   H₂O

4                          5

4                          4                     4                   4

0                          1                      1                    1

Concentration of [OH⁻] from NaOH :

\tt \dfrac{1~mlmol}{20+25~ml}=0.02

pOH=-log[OH⁻]

pOH=-log 10⁻²=2

pH+pOH=14

pH=14-2=12

3 0
3 years ago
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