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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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Bromine has a density of 3.10g/cm3. If you have 50.0 ML of bromine, how many grams do you have?
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Answer:

The answer is

<h2>155 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of bromine = 50 mL

density = 3.10 g/cm³

It's mass is

mass = 50 × 3.10

We have the final answer as

<h3>155 g</h3>

Hope this<u> </u>helps you

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(I)how many atoms are present in 7g of lithium?
ICE Princess25 [194]

Answer :

(i) The number of atoms present in 7 g of lithium are, 6.07\times 10^{23}

(ii) The number of atoms present in 7 g of lithium are, 1.204\times 10^{24}

(iii) The number of moles of F_2 is, 1 mole

The number of moles of CO_2 is, 0.5 mole

The number of moles of OH^- is, 1 mole

Explanation :

<u>Part (i) :</u>

First we have to calculate the moles of lithium.

\text{Moles of }Li=\frac{\text{Mass of }Li}{\text{Molar mass of }Li}

Molar mass of Li = 6.94 g/mole

\text{Moles of }Li=\frac{7g}{6.94g/mol}=1.008mole

Now we have to calculate the number of atoms present.

As, 1 mole of lithium contains 6.022\times 10^{23} number of atoms

So, 1.008 mole of lithium contains 1.008\times 6.022\times 10^{23}=6.07\times 10^{23} number of atoms

Thus, the number of atoms present in 7 g of lithium are, 6.07\times 10^{23}

<u>Part (ii) :</u>

First we have to calculate the moles of carbon.

\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}

Molar mass of C = 12 g/mole

\text{Moles of }C=\frac{24g}{12g/mol}=2mole

Now we have to calculate the number of atoms present.

As, 1 mole of carbon contains 6.022\times 10^{23} number of atoms

So, 2 mole of carbon contains 2\times 6.022\times 10^{23}=1.204\times 10^{24} number of atoms

Thus, the number of atoms present in 7 g of lithium are, 1.204\times 10^{24}

<u>Part (iii) :</u>

<u>To calculate the moles of </u>F_2<u> :</u>

\text{Moles of }F_2=\frac{\text{Mass of }F_2}{\text{Molar mass of }F_2}

Molar mass of F_2 = 38 g/mole

\text{Moles of }F_2=\frac{19g}{19g/mol}=1mole

Thus, the number of moles of F_2 is, 1 mole

<u>To calculate the moles of </u>CO_2<u> :</u>

\text{Moles of }CO_2=\frac{\text{Mass of }CO_2}{\text{Molar mass of }CO_2}

Molar mass of CO_2 = 44 g/mole

\text{Moles of }CO_2=\frac{22g}{44g/mol}=0.5mole

Thus, the number of moles of CO_2 is, 0.5 mole

<u>To calculate the moles of </u>OH^-<u> ions :</u>

\text{Moles of }OH^-=\frac{\text{Mass of }OH^-}{\text{Molar mass of }OH^-}

Molar mass of OH^- = 17 g/mole

\text{Moles of }OH^-=\frac{17g}{17g/mol}=1mole

Thus, the number of moles of OH^- is, 1 mole

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3 years ago
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