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3 years ago

Describe two negative effects that water pollution can have on the environment. Explain one way water pollution can be reduced.

2 answers:

7 0

Water pollution affects the fish, and then we eat the fish so the fish affects us. im not sure about the first part.Do not dispose of household cleaning down the toilet.

Romashka-Z-Leto [24]3 years ago

5 0

Answer:

1) Adverse effect on aquatic life also increase in biological oxygen demand.

2) Biomagnification of pollutants

Explanation:

Water pollution is caused by unwanted addition of waste, animal debris, industrial effluents etc in a water body.

Water pollution may cause many direct and indirect harmful effects on environment. To name a few:

a) Decrease in level of oxygen leading to death of aquatic organism like fish.

Increase in biological oxygen demand.

b) Biomagnification: a pollutant may enter food chain from water and some pollutant may enter food chain contaminating other members of food chain.

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CH4 + 202 → CO2 + 2H2O How many grams of O2 needed to produce 36 grams of H2O?

Kipish [7]

<h3>Answer:</h3>

64 g O₂

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 36 g H₂O

[Solve] x g O₂

Step 2: Identify Conversions

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Stoichiometry

Set up conversion: $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
Divide/Multiply [Cancel Units]: $\displaystyle 63.929 \ g \ O_2$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

63.929 g O₂ ≈ 64 g O₂

8 0

3 years ago

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Alekssandra [29.7K]

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2 years ago

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STALIN [3.7K]

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4 0

3 years ago

How would you prepare 3.5 L of a 0.9M solution of KCl?

Fudgin [204]

Calculate the mass of the solute in the solution :

Molar mass KCl = 74.55 g/mol

m = Molarity * molar mass * volume

m = 0.9 * 74.55 * 3.5

m = 234.8325 g

To prepare 0.9 M KCl solution, weigh 234.8325 g of salt in an analytical balance, dissolve in a beaker, shortly after transfer with the help of a funnel of transfer to a volumetric flask of 100 cm³ and complete with water up to the mark, then cover the balloon and finally shake the solution to mix

hope this helps!

8 0

3 years ago

A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is

Romashka [77]

Answer:

$\large \boxed{\text{B.) 2.8 atm}}$

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ = 1520 Torr; T₁ = 27 °C

p₂ = ?; T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\$

(c) Convert the pressure to atmospheres

$p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}$

7 0

2 years ago