Effect of increasing surface area on the rate of a reaction. ... Increasing the surface area of a solid reactant exposes more of its particles to attack. This results in an increased chance of collisions between reactant particles, so there are more collisions in any given time and the rate of reaction increases.
Technically there is only one phase unless you account for a solution where you have a pure liquid with something dissolved in it. Unless you count aqueous as a phase which is just dissolved. Since you are in high school the answer you are looking for is one. <span />
Nume smechere )(∂ⓓᗴηⓛ⫸⫷nume smechere )(∂ⓓᗴηⓛ⫸⫷nume smechere )(∂ⓓᗴηⓛ⫸⫷nume smechere )(∂ⓓᗴηⓛ⫸⫷nume smechere )(∂ⓓᗴηⓛ⫸⫷nume smechere )(∂ⓓᗴηⓛ⫸⫷nume smechere )(∂ⓓᗴηⓛ⫸⫷nume smechere )(∂ⓓᗴηⓛ⫸⫷nume smechere )(∂ⓓᗴηⓛ⫸⫷
<u>Answer:</u> The new pressure will be 101.46 kPa.
<u>Explanation:</u>
To calculate the new pressure, we use the equation given by Gay-Lussac Law. This law states that pressure is directly proportional to the temperature of the gas at constant volume.
The equation given by this law is:
where,
are initial pressure and temperature.
are final pressure and temperature.
We are given:
By using conversion factor: 
Putting values in above equation, we get:
Hence, the new pressure will be 101.46 kPa.
boron, carbon, nitrogen, oxygen and flourine families in addition to the noble gases.
_________________________________________________________
also to answer the 'hello guys what is good!':nothin u?
__________________________________________
o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0
