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lapo4ka [179]
3 years ago
5

What is the shape of graphite​

Chemistry
1 answer:
VashaNatasha [74]3 years ago
7 0

Answer:

The molecular formula lists the symbol of each element within the compound followed by a number (usually in subscript). The letter and number indicate how many of each type of element are in the compound. If there is only one atom of a particular element, then no number is written after the element.

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What is the density of an object having a mass of 16 g and a volume of 30 mL ?
Marizza181 [45]

Answer:

12

Explanation:

7 0
3 years ago
Read 2 more answers
In which of the following situations is the greatest amount of work accomplished?
Dahasolnce [82]

<span>work = force x distances </span>

<span>A - moving 2 newton's up 0.6 meter = 1.2 joules
B - moving 4 newton's up 0.6 meter = 2.4 joules
C - moving 6 newton's up 0.3 meter = 1.8 joules
D - moving 9 newton's up 0.3 meter = 2.7 joules
The greatest amount of work is in example D. 
</span>

<span>
 D is your answer. </span>

7 0
3 years ago
Read 2 more answers
Plz help me thank u will give brainliest
Triss [41]
I believe it is b. you only want to change one thing at a time so you know which one thing caused the effect
7 0
3 years ago
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
Mass=1.01 kg; volume =1000cm3
guajiro [1.7K]

Density = 1.01 g/cm^3 or 1.01 kg/dm^3 or 1010 kg/m^3

Density = mass/volume = 1010 g/1000 cm^3 = 1.01 g/cm^3 = 1.01 kg/dm^3

= 1010 kg/m^3


7 0
3 years ago
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