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3 years ago

Can someone give me a simple example of a exothermic reaction? Would a campfire be one?

Chemistry

1 answer:

Nitella [24]3 years ago

5 0

4 Examples of Exothermic Reaction:

Formation of snow
Burning wood
Burning Candle
Gas burner in use

- and yes, a campfire can be a exothermic reaction.

$\color{A689E1}I$ $\color{A689E1}hope$ $\color{A689E1}this$ $\color{A689E1}helps!$ ^-^

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A water treatment plant applies chlorine for disinfection so that 10 mg/L chlorine is achieved immediately after mixing. The vol

Vlada [557]

<u>Answer:</u> The mass of chlorine needed by the plant per day is $9.4625\times 10^8mg$

<u>Explanation:</u>

We are given:

Volume o water treated per day = 25,000,000 gallons

Converting this volume from gallons to liters, we use the conversion factor:

1 gallon = 3.785 L

So, $\frac{3.785L}{1\text{ gallon}}\times 25,000,000\text{ gallons}=9.4625\times 10^7L$

Amount of chlorine applied for disinfection = 10 mg/L

Applying unitary method:

For 1 L of water, the amount of chlorine applied is 10 mg

So, for $9.4625\times 10^7L$ of water, the amount of chlorine applied will be $\frac{10mg}{1L\times 9.4625\times 10^7L}=9.4625\times 10^8mg$

Hence, the mass of chlorine needed by the plant per day is $9.4625\times 10^8mg$

6 0

3 years ago

Consider a cell membrane at 298K which has glucose at the concentration of 200mM inside the cell and 20 mM outside the cell, wit

OleMash [197]

Answer:

D. +5.7 kJ/mol

Explanation:

Molar free energy (ΔG) in the transportation of uncharged molecules as glucse through a cell membrane from the exterior to the interior of the cell is defined as:

ΔG = RT ln C in / C out

knowing R is 8,314472 kJ/molK; T is 298K Cin = 200mM and Cout = 20mM

ΔG = 5,7 kJ/mol

Right answer is:

D. +5.7 kJ/mol

I hope it helps!

4 0

3 years ago

1.42 g H2 is allowed to react with 10.4 g N2 , producing 2.14 g NH3 . Part A What is the theoretical yield in grams for this rea

Bad White [126]

Taking into account the reaction stoichiometry, the theorical yield for the reaction is 8.0467 grams of NH₃.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 H₂ + N₂ → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

H₂: 3 moles
N₂: 1 mole
NH₃: 2 moles

The molar mass of the compounds is:

H₂: 2 g/mole
N₂: 28 g/mole
NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

H₂: 3 moles ×2 g/mole= 6 grams
N₂: 1 mole ×28 g/mole= 28 grams
NH₃: 2 moles ×17 g/mole= 34 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 28 grams of N₂ reacts with 6 grams of H₂, 10.4 grams of N₂ reacts with how much mass of H₂?

$mass of H_{2} =\frac{10.4 grams of N_{2}x 6 grams of H_{2} }{28 grams of N_{2}}$

<u><em>mass of H₂= 2.2286 grams</em></u>

But 2.2286 grams of H₂ are not available, 1.42 grams are available. Since you have less mass than you need to react with 10.4 grams of N₂, H₂ will be the limiting reagent.

<h3>Definition of theorical yield</h3>

The theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Theoretical yield in this case</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 34 grams of NH₃, 1.42 grams of H₂ form how much mass of NH₃?

$mass of NH_{3} =\frac{1.42 grams of H_{2} x 34 grams of NH_{3}}{6grams of H_{2} }$