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erik [133]
3 years ago
11

North Carolina is a rapidly urbanizing state. This growth increases the number of homes, roads, and the need for goods and servi

ces. All of this can be a great boost to the local economy.
How could this growth result in a negative impact on the environment?
A) An increase in biodiversity.
B) An increase in the air purity.
C) An increase in the water quality.
D) An increase in the amount of waste.
Chemistry
1 answer:
Serjik [45]3 years ago
7 0
I think it's D) An increase in the amount of waste.
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The highest parts of a transverse wave are called a. troughs. b. crests. c. nodes. d. wavelengths.
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The highest is b crest.
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3 years ago
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The movement of water both on and<br> above Earths surface
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4 0
3 years ago
When 551. mg of a certain molecular compound X are dissolved in 100 g of benzonitrile (CH,CN), the freezing point of the solutio
arsen [322]

Answer:

1.12g/mol

Explanation:

The freezing point depression of a solvent for the addition of a solute follows the equation:

ΔT = Kf*m*i

<em>Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)</em>

<em>ΔT = 13.4°C - (-12.82) = 26.22°C</em>

<em>m is molality of the solution</em>

<em>Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)</em>

<em>And i is Van't Hoff factor (1 for all solutes in benzonitrile)</em>

Replacing:

26.22°C = 5.35°Ckgmol⁻¹*m*1

4.90mol/kg = molality of the compound X

As the mass of the solvent is 100g = 0.100kg:

4.9mol/kg * 0.100kg = 0.490moles

There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:

0.551g / 0.490mol

= 1.12g/mol

<em>This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.</em>

5 0
2 years ago
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for
inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

Cu^{2+}(aq)+Fe(s)\rightarrow Cu(s)+Fe^{2+}(aq)

Reduction at cathode :

Cu^+(aq)+2e^-\rightarrow Cu(s)

Reduction potential of  Cu^{2+} to Cu=E^o_{1}=0.34 V

Oxidation at anode:

Fe(s)\rightarrow Fe^{2+}(aq)+2e^-

Reduction potential of  Fe^{2+} to Fe=E^o_{2}=-0.44 V

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}

Putting values in above equation, we get:

E^o_{cell}=0.34V -(-0.44 V)=0.78 V

The standard cell potential of the reaction is 0.78 Volts.

3 0
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Answer:

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