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2 years ago

Pesticides are best described as _______.

Chemistry

2 answers:

Aleksandr-060686 [28]2 years ago

5 0

D. toxic chemical used to control pest population.

OlgaM077 [116]2 years ago

3 0

Pesticides are best described as D. toxic chemicals used to control pest populations.

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That's just the tip of the iceberg" is a popular expression you may have heard. It means that what you can see is only a small p

puteri [66]

Answer:

B 1.23 g/cc

Explanation:
For something to float on seawater, the density must be less than 1.03 g/mL. If the object sinks, the density is greater than 1.03 g/mL.

Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.

A. 0.88 g/cc
This is less than 1.03 g/cc, which would result in floating.

B. 1.23 g/cc
This is the best answer choice. The iceberg is mostly beneath the water, but some of it is exposed. The density is greater than 1.03 g/mL, but not so much greater that it would immediately sink.

C. 0.23 g/cc
This is less than 1.03 g/cc, which would produce floating.

D. 4.14 g/cc
This is much greater than 1.03 g/cc and the result would be sinking.

5 0

2 years ago

Automobiles are often implicated as contributors to global warming because they are a source of the greenhouse gas CO2. How many

kondor19780726 [428]

Answer:

5,747.82 pounds of $CO_2$ would your car release in a year if it was driven 170 miles per week.

Explanation:

Mileage of the car = 27.5 miles/gal

Distance driven by car in week = 170 miles

Distance driven by car in a day= $\frac{170 miles}{7}=24.286 mile$

Volume of gasoline used in a day : V

$\frac{24.286 mile}{27.5 miles/gal}=0.8831 gal=0.8831\times 3785.41 cm^3=3,342.96 cm^3$

$1 gal = 3785.41 mL = 3785.41 cm^3$

Density of the gasoline = d= $0.692 g/cm^3$

Mass of the gasoline used in a day = m

$m=d\times V=0.692 g/cm^3\times 3,342.96 cm^3=2,313.33 g$

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

Moles of gasoline ( isooctane):

= $\frac{2,313.33 g}{114 g/mol}=20.292 mol$

According to reaction , 2 moles of isooctane gives 16 moles of carbon dioxide gas.Then 20.292 mole isooctane will give :

$\frac{16}{2}\times 20.292 mol=162.340 mol$ of carbon dioxide gas

Mass of 162.340 moles of carbon dioxide gas :

162.340 mol × 44 g/mol = 7,142.91 g

Mass of carbon dioxide gas produced in a day = 7,142.91 g

1 year = 365 days

Mass of carbon dioxide gas produced in a 365 days:

= 365 × 7,142.91 g = 2,607,163.494 g

1 pound = 453.592 g

$2,607,163.494 g=\frac{ 2,607,163.494}{453.592} pounds=5,747.82 pounds$

5,747.82 pounds of $CO_2$ would your car release in a year if it was driven 170 miles per week.

7 0

2 years ago

Which of the following does not participate in, nor is a component of, the electron transport chain? (Remember that molecules ca

Yanka [14]

Answer:

A) coenzyme A

Explanation:

The NADH and FADH₂ are the energy rich molecules which are formed in the processes like glycolysis, TCA cycle and the fatty acid oxidation as they contain pair of electrons which have very high transfer potential.

As a result of the energy produced when these molecules transfer their electrons to the oxygen , ATP is generated by a series of electron carriers which collectively is called electron transport chain (ETC).

The components of chain include Fe–S centers, Non-heme, FMN, coenzyme Q, and cytochromes . 

The energy derived from the transfer of electrons is used to pump the protons across mitochondrial membrane.

As a result, an electrochemical gradient is generated which results in some energy which is then harnessed by the ATP synthase to form ATP.

6 0

1 year ago

Consider the air over a city to be a box 100km on a side that reaches up to an altitude of 1.0 km. Clean air is blowing into the

Mariana [72]

Explanation:

It is known that equation for steady state concentration is as follows.

$C_{a} = \frac{QC}{Q + kV}$

where, Q = flow rate

k = rate constant

V = volume

C = concentration of the entering air

Formula for volume of the box is as follows.

V = $a^{2}h$

= $100 \times 100 \times 1$

= $10000 km^{3}$

Now, expression to determine the discharge is as follows.

Q = Av

= $100 \times 1 \times \frac{4 m}{s} \times \frac{km}{1000 m}$

= 0.4 $km^{3}/s$

And, m (loading) = 10kg/s,

k = 0.20/hr

as 1 $km^3 = 10^{12}$ L (if u want kg/L as concentration)

Now, calculate the concentration present inside as follows.

$C_{in} = \frac{10kg/s}{0.4 km^3/s}$

= 25 $kg/km^3$

Now, we will calculate the concentration present outwards as follows.

$C_{out} = {C_{in}}{(1 + k \times t)}$ ,

and, t = $\frac{V}{Q}$

= 25000 s or 6.94 hr

Hence, $C_{out} = \frac{25}{(1 + 0.20 \times 6.94)}$

= 10.47 $kg/km^3$

Thus, we can conclude that the the steady-state concentration if the air is assumed to be completely mixed is $C_{out} = 10.47 kg/km^3$ and $C_{in} = 25 kg/km^3$ .

4 0

1 year ago

Complete the following radioactive decay problem. 210/84 Po --> 206/82 Pb + ____/____ __

vazorg [7]

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

Your equation is:

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + ?

It becomes easier to balance the equation if we replace the "?" with an element symbol _x^yZ.

Then the equation becomes

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb +_x^yZ

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Sum of superscripts: 210 = 206 + y, so y =4.

Sum of subscripts: 84 = 82 + x, so x = 2.

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

We should recall that $_2^4He$ is an α particle.

Thus, the equation represents the α decay of polonium-210 to lead-206.

The nuclear equation is

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

4 0

2 years ago