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2 years ago

Explain why some pesticides bioaccumulate whereas others do not

1 answer:

7 0

Bioaccumulation refers to the accumulation of chemicals in a living organism. The compound or chemical accumulates at a rate faster than it is being metabolized or excreted by the organism. Chemicals bioaccumulate by binding to the proteins and fats in an organism while others bioaccumulate through the repeated consumption of contaminated organisms.

Pesticides containing chemicals that dissolve easily in fat but not in water tend to bioaccumulate. Pesticides that contain chemicals that can easily be metabolized by organisms do not bioaccumulate. In summary, the nature of the chemical used in pesticides and the capability of organisms to metabolize the said chemicals can dictate whether it will bioaccumulate or not.

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What are the characteristics of a virtual image? Check all that apply.

tia_tia [17]

<h3>Answer:</h3>

Can only be seen while looking through a lens
May appear larger than the object
Appears to come from behind the lens

<h3>Explanation:</h3>

To answer this question we need to ask ourselves;

What is a virtual image?

A virtual image is an image that can not be formed on a screen.
It is formed by rays that appear to converge at a point.

What are the characteristics of a virtual image?

A virtual image can only be seen while looking through lenses
The image appear to come from behind the lens
It may be larger than the object
It can not be projected onto a screen.
It is always upright

6 0

2 years ago

Read 2 more answers

6. This particle has no charge protons neutrons electrons

Stella [2.4K]

Answer:

neutron, has no charge

5 0

3 years ago

Read 2 more answers

What happens if you don't disconnect the wires in a series circuit?

Mars2501 [29]

Answer:

the series circuit has one current so if you don't disconnect the wires nothing happens

but if you disconnect the wires in series circuit, the current stop and the current doesn't go to the end

Explanation:

(≧▽≦)hope it helps (≧▽≦)

5 0

2 years ago

Calculate the volume of each of the following gases at STP

kherson [118]

For Ar :

1 mol ------------ 22.4 L ( at STP )
7.6 mol ---------- x L

x = 7.6 * 22.4

x = 170.24 L
-----------------------------------------------------------------
For C2H3:

1 mol ------------ 22.4 ( at STP)
0.44 mol --------- y L

y = 0.44 * 22.4

y = 9.856 L

hope this helps !.

7 0

3 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

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

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

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

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

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

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

$P_1V_1=P_2V_2$

3. Solution

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

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

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.

3. Solution

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

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

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

3. Solution

Multiply 865 mmHg by the conversion factor:

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

3 0

3 years ago