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2 years ago

8

Fill in the blank. _______rain falls in areas that have air pollution.

1 answer:

Genrish500 [490]2 years ago

7 0

Acid rain falls in areas that have air pollution. (<span>Acid rain is caused by emissions of </span>sulfur dioxide<span> and </span>nitrogen oxide)

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How/why is maintaining healthy soil a social justice issue for our area and community

podryga [215]

Answer:

Social justice aims to give individuals and groups fair treatment and an impartial share of social, environmental and economic benefits. The concept promotes the fair distribution of advantages and disadvantages within a society, regardless of background and status.

Environmental justice deals explicitly with the distribution of environmental benefits and the burdens people experience, at home, at work, or where they learn, play and spend leisure time. Environmental benefits include attractive and extensive greenspace, clean air and water, and investment in pollution abatement and landscape improvements. Environmental burdens include risks and hazards from industrial, transport-generated and municipal pollution.

3 0

2 years ago

: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne

olchik [2.2K]

Answer:

The Net reaction is

$H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction$
$H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH$
$HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH$

Explanation:

From the Question we are told that the buffers are

$H_2C_6H_5O_7^ -$ and $HC_6H_5O_7 ^{ 2-}$

When NaOH is added the Net ionic reaction would be

$H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction$
$H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH$
$HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH$

8 0

3 years ago

Read 2 more answers

In laboratory experiment, a NOVDEC Student was

Ede4ka [16]

Answer:

i. Molar mass of glucose = 180 g/mol

ii. Amount of glucose = 0.5 mole

Explanation:

<em>The volume of the glucose solution to be prepared</em> = 500 $cm^3$

<em>Molarity of the glucose solution to be prepared</em> = 1 M

i. Molar mass of glucose ( $C_1_2H_6O_6$ ) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g/mol

ii.<em> mole = molarity x volume</em>. Hence;

amount (in moles) of the glucose solution to be prepared

= 1 x 500/1000 = 0.5 mole

3 0

2 years ago

At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration

wlad13 [49]

Answer:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.03901 mol/liter

[Cl₂O] = 0.02351 mol/liter

Explanation:

<u />

<u>1. Chemical reaction:</u>

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

<u>2. Initial concentrations:</u>

i) 1.3 g H₂O

Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol

Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol

Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

<u>3. ICE (Initial, Change, Equilibrium) table</u>

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

I 0.0481 0.0326 0

C -x -x +x

E 0.0481-x 0.0326-x x

<u />

<u>4. Equilibrium expression</u>

$K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}$

$0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}$

<u />

<u>5. Solve:</u>

$x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0$

Use the quadatic formula:

$x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}$

The positive result is x = 0.00909

Thus the concentrations are:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter

[Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

3 0

3 years ago

Water is composed of two hydrogen atoms and one oxygen atom therefore, water is a

vladimir1956 [14]

two hydrogen
compound !!

4 0

3 years ago