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3 years ago

5

Describe what you can do to reduce non-point source water pollution. List at least 3 specific actions, or, write 3 sentences abo

ut 1 action.

1 answer:

vovangra [49]3 years ago

4 0

Grass planting, laying of straw , putting up sediment fenced or knee high black fabric fences can help reduce no point source water pollution

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Which is the most likely to be reduced?

NISA [10]

Answer : The correct option is, $Zn^{2+}$

Explanation :

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. That means, the loss of electrons takes place.

Or we can say that, oxidation reaction occurs when a reactant losses electrons in the reaction.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. That means, the gain of electrons takes place.

Or we can say that, reduction reaction occurs when a reactant gains electrons in the reaction.

According to the electrochemical series, $Zn^{2+}$ most likely to be reduced because

Hence, the ion most likely to be reduced is $Zn^{2+}$ .

3 0

3 years ago

Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and

kompoz [17]

Answer:

a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)

b. i. NO₂⁻ is the oxidizing agent

ii. NO₃⁻ is the reducing agent.

Explanation:

a. Balance the following skeleton reaction

The reaction is

NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)

The half reactions are

NO₂ (g) → NO₃⁻ (aq) (1) and

NO₂ (g) → NO₂⁻ (aq) (2)

We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)

We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms

NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)

The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)

Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.

So, NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)

We now add equation (4) and (5)

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)

+ NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)

2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + e⁻ (4)

2NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq)

We now add two hydroxide ions to both sides of the equation.

So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + 2OH⁻ (aq)

The hydrogen ion and the hydroxide ion become a water molecule

2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H₂O (l)

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)

So, the required reaction is

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)

b. Identify the oxidizing agent and reducing agent

Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = 0

x - 4 = 0

x = 4

Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = -1

x - 4 = -1

x = 4 - 1

x = 3

Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = -1

x + 3(-2) = -1

x - 6 = -1

x = 6 - 1

x = 5

i. The oxidizing agent

The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus NO₂⁻ is the oxidizing agent

ii. The reducing agent

The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and NO₃⁻ is the reducing agent.

3 0

3 years ago

A 12 gram piece of metal is heated to 300 °C from 100 °C with 1120 Joules of energy. What is the specific heat of the metal?

Sergeeva-Olga [200]

Answer:

The specific heat for the metal is 0.466 J/g°C.

Explanation:

Given,

Q = 1120 Joules

mass = 12 grams

T₁ = 100°C

T₂ = 300°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

ΔT = T₂ - T₁ = 300°C - 100°C = 200°C

Substituting values,

1120 = (12)(200)(Cp)

Cp = 0.466 J/g°C.

Therefore, specific heat of the metal is 0.466 J/g°C.

7 0

3 years ago

a lead ball has a mass of 5.910 g and a diameter or 10.0 mm what is the density of lead in grams per cubic centimeters .The volu

storchak [24]

Answer:

11.3 g/cm³

Explanation:

Data Given:

Diameter of lead ball = 10.0 mm

Convert mm to cubic centimeter (cm³)

1 mm = 0.1 cm

So,

10.0 mm = 10.0 x 0.1 = 1 cm

mass of lead ball = 5.910 g

density of of lead ball =?

Solution:

To find density formula will be used

d = m/V ...........(1)

where

d = density

m = mass

V = volume

So,

To find the density of the lead ball (sphere), first we have to find volume of the ball.

To find volume of the ball following formula will be used

V = 4 . π . r³/ 3 ..............(2)

where

π = 3.14

r = radius

So,

r = D/2

as

D = 1 cm

So, radius will be

r = 1 cm / 2

r = 0.5 cm

Now put values in formula 2

V = (4)(3.14)(0.5cm)³/ 3

V = (12.56)(0.125 cm³) /3

V = 1.57 cm³/3

V = 0.523 cm³

So, the volume of the lead ball is 0.523 cm³

Now put values in equation 1

d = m/V

d = 5.910 g / 0.523 cm³

d = 11.3 g/cm³

Density of the lead ball = 11.3 g/cm³

7 0

4 years ago

Can somebody please help me with these I literally have no idea what it’s asking

noname [10]

Answer:

I believe it's copper

Explanation:

The reasoning being that Copper = 9.0 g/cm and the mystery item has 5.0 cm so 9*5 = 45 something like that

6 0

3 years ago