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3 years ago

What changed about the peppered moth population during the second industrial revolution

Chemistry

2 answers:

Marysya12 [62]3 years ago

7 0

Answer:

The months became darker colored and more darker colored moths were sighted.

Explanation:

All of the new air pollution caused the moths to alter there wings and bodies. The moths had no adaptation to this air pollution, causing them to be discolored.

Eduardwww [97]3 years ago

7 0

Answer:

The color of the moths

Explanation:

<em>It izz wat it izzzz!!!</em>

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Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6

Sergeeva-Olga [200]

Explanation:

The given data is as follows.

Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml

Therefore, molarity of toulene is calculated as follows.

Molarity = $\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}$

= 4.11 M

Hence, molarity of toulene is 4.11 M.

As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

Molality = $\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}$

= 8.6 m

Hence, molality of given toulene solution is 8.6 m.

Now, calculate the number of moles of toulene as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{75.8 g}{92.13 g/mol}$

= 0.8227 mol

Now, no. of moles of benzene will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{95.6 g}{78.11 g/mol}$

= 1.2239 mol

Hence, the mole fraction of toulene is as follows.

Mole fraction = $\frac{\text{moles of toulene}}{\text{total moles}}$

= $\frac{0.8227 mol}{(0.8227 + 1.2239) mol}$

= 0.402

Hence, mole fraction of toulene is 0.402.

As density of given solution is 0.857 $g/cm^{3}$ so, we will calculate the mass of solution as follows.

Density = $\frac{mass}{volume}$

0.857 $g/cm^{3}$ = $\frac{mass}{200 ml}$ (As 1 $cm^{3}$ = 1 g)

mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

Mass % = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

= $\frac{75.8 g}{171.4 g} \times 100$

= 44.22%

Therefore, mass percent of toulene is 44.22%.

8 0

3 years ago

Which of the following could not happen to an igneous rock?

galben [10]

I could not break unless you hit it with a sledge hammer

4 0

2 years ago

Which of the following is not an effect of acid rain? a. Corrosion of buildings c. Limestone corrosion b. Reduced cold tolerance

Novay_Z [31]

<span> d. Reduced visibility

Acid rain doesn't reduce visibility, smog does
</span>

3 0

3 years ago

Cathode-electrolyte interphase in lithium batteries revealed by cryogenic electron microscopy. matte

Ksju [112]

Cryo-EM is used to preserve and characterize cycled positive electrodes. Under regular cycling conditions, there isn't an intimate coating layer like CEI.A small electrical short can cause a stable conformal CEI to form in place. The conformal CEI's chemistry is revealed by EELS and cryo-(S)TEM.

It has been assumed that the intimate coating layer generated on the positive electrode, known as cathode electrolyte interphase (CEI), is crucial. However, there are still numerous questions about CEI. This results from the absence of useful instruments to evaluate the chemical and structural characteristics of these delicate interphases at the nanoscale. Here, using cryogenic electron microscopy, we establish a methodology to maintain the natural condition and directly see the interface on the positive electrode.

Learn more about Cathode electrolyte interphase here:

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4 0

1 year ago

Calculate the concentration in % (m/m) of a solution containing 30.0g of mgcl2 dissolved in 270.0g of h20

qaws [65]

As the question tells you, you need to use the formula

% mass= mass of solute/ mass of solution x 100

mass solute= 30.0 g
mass of solution= 30.0 + 270.0= 300.0 g

% mass= 30.0/ 300.0 x 100= 10%

answer is B

4 0

2 years ago