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damaskus [11]
2 years ago
10

How is dumping our waste different from dilution and dispersion?

Chemistry
1 answer:
ch4aika [34]2 years ago
6 0
Dumping waste is different from dilution and dispersion. Dumping waste is directly disposing your waste and not regulating its effect to the environment. Dilution is done usually before waste water disposal, meaning adding water to the waste to minimize its concentration. 
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In the process of electrolysis current can flow through a liquid because
Harman [31]
Negative ions are attracted to the anode. In the process of electrolysis, current can flow through a liquid because negative ions are attracted to the anode.
3 0
2 years ago
Read 2 more answers
Consider the balanced equation below. 4nh3 3o2 right arrow. 2n2 6h2o what is the mole ratio of nh3 to n2? 2:4 4:2 4:4 7:2
KIM [24]

The required mole ratio of  NH₃ to N₂ in the given chemical reaction is 2:4.

<h3>What is the stoichiometry?</h3>

Stoichiometry of the reaction gives idea about the number of entities present on the reaction before and after the reaction.

Given chemical reaction is:

4NH₃ + 3O₂ → 2N₂ + 6H₂O

From the stoichiometry of the reaction it is clear that:

4 moles of NH₃ = produces 2 moles of N₂

Mole ratio NH₃ to N₂ is 2:4.

Hence required mole ratio is 2:4.

To know more about mole ratio, visit the below link:
brainly.com/question/504601

3 0
2 years ago
I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

3 0
3 years ago
Jose times how long sugar takes to dissolve in warm water. he conducts four trials of his experiment. what should he conclude fr
mylen [45]

He can conclude that his experiment has very low precision.

<h3>What is Precision ?</h3>

Precision is defined as the degree of refinement with which an operation is performed or a measurement .

Precision is how close the exact answers are together.

As, the answers are increasing in time.

None of the answers are similar to one another.

Hence, He can conclude that his experiment has very low precision.

Learn more about Precision here ;

brainly.com/question/27845433

#SPJ1

7 0
2 years ago
2NOCl(g) ↔ 2NO(g) + Cl2(g) with K = 1.6 x10–5. In an experiment, 1.00 mole of pure NOCl and 1.00 mole of pure Cl2 are placed in
poizon [28]

Answer: 3.8\times 10^{-3}M

Explanation:

Moles of  NOCl = 1 mole

Moles of  Cl_2 = 1 mole

Volume of solution = 1 L

Initial concentration of NOCl = 1 M

Initial concentration of Cl_2 = 1 M

The given balanced equilibrium reaction is,

                  2NOCl(g)\rightleftharpoons 2NO(g)+Cl_2(g)

Initial conc.          1 M                      0M          1 M

At eqm. conc.     (1-2x) M              (2x) M       (1+x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[NO]^2[Cl_2]}{[NOCl]^2}

The K_c= 1.6\times 10^{-5}}

Now put all the given values in this expression, we get :

{1.6\times 10^{-5}}=\frac{(2x)^2\times (1+x)}{(1-2x)^2}

By solving the term 'x', we get :

x=0.0019

Concentration of NO at equilibrium= (2x) M  =  2\times 0.0019=3.8\times 10^{-3}M

6 0
2 years ago
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