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Gnoma [55]
3 years ago
6

A catalyst is used to increase the reaction rate of a chemical reaction.

Chemistry
1 answer:
hichkok12 [17]3 years ago
3 0
The answer would be C
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10. pls pls help me
vagabundo [1.1K]

Answer:

2c2h2+502=4co2+H2O

Explanation:

this is the answer

8 0
3 years ago
As an electron gains energy, it becomes excited and jumps to a lower energy level?
Maru [420]

Yes if you add an energy to an electron the electron will become excited, and it will jump to its highest level then go back down releasing  energy


3 0
3 years ago
Read 2 more answers
The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consis
Dimas [21]
The solution is as follows:

K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106

The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol

Use the ICE approach

        borneol  →  isoborneol
I         0.0972           0.0486
C         -x                     +x
E     0.0972 - x        0.0486 + x

Total moles = 0.1458

Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P

0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832 

Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
6 0
3 years ago
Which element does this Bohr model represent? (look at a periodic table if needed)
Serjik [45]
Circulating round the nucleus are the electrons in various orbits of different energy levels. Electrons are negatively charged and represented by the symbol 'e'. In the given image the number of protons are -6. Hence the element in question is Carbon as Carbon has the atomic number 6.
6 0
2 years ago
CAN SOME ONE HELLLLLLLLLLLLLLLLLLLLLLP WILL GIVE BRAINLIEST TO THE BEST ANSWR FOR THIS AND GIVE AWAY 15 POINTS
lord [1]
1. C
2. C
3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit. 
4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust.  Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress. 

I hope this helped! These are COMPLEX questions though! =D
8 0
3 years ago
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