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ICE Princess25 [194]
3 years ago
6

At STP, solid carbon can exist as diamond and graphite. Compared to the molecular structure and chemical properties of diamond,

graphite has
(1) a different molecular structure and different properties
(2) a different molecular structure and the same properties
(3) the same molecular structure and different properties
(4) the same molecular structure and the same properties
Chemistry
2 answers:
7nadin3 [17]3 years ago
5 0

Answer is: (1) a different molecular structure and different properties.

Different forms of the same element that have different properties because of different atom arrangements are called allotropes.

Carbon has many allotropes, but two most important are graphite and diamomd.

Graphite has sp2 and diamond has sp3 hybridization of carbon atoms, because of that graphite conduct electricity and diamond not.

In diamond carbon atoms are arranged in the face centered cubic crystal structure called a diamond lattice.  

Diamond has very strong covalent bonds between carbon atoms and because of that it has the highest hardness and thermal conductivity of any bulk substance.

il63 [147K]3 years ago
5 0
The answer is (1) a different molecular structure and different properties. The graphite has a molecular structure of layer and in the diamond, each atom are bonded strongly to four adjacent carbon atoms which formed a three-sided pyramid. 
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5 0
4 years ago
Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int
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Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

Million Gallons per day 1 MGD = 3785411.8 litre/day = 3785411.8 L/d

F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d

We have one flow of wastewater released into a stream.  

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

C_f = \frac{F1*C1 +F2*C2}{F1 +F2}

Replacing every value in L/d and mg/L

C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).  

We have the total flow F3 = F1 + F2 and the final concentration C_f. It is required to calculate per day, let's take a time of t = 1 day.  

F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg

After that, mg are converted to pounds.  

M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb

b) A total of 137.6 pounds pass a given spot downstream per day.

4 0
4 years ago
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