The same
Explanation:
If a liquid substance is transferred to a different container, the volume of the liquid in the new container will remain the same.
The volumes of liquids are fixed and does not change. Wherever they are contained, just like solids, they maintain their constant space.
- Volume is the amount of space occupied by a body.
- Gases do not have fixed volume as they fill their containers and they take up the shape.
- Solids and liquids have a fixed volume.
- They do not change their volume.
learn more:
State of matter brainly.com/question/10972073
#learnwithBrainly
Proton number = the atomic number (which is the smaller number
neutron number = the mass number (the bigger number) - the atomic number
number of electrons = the atomic number - the charge (it depends on the element but group 1 is +1 group 2 is +2 group 3 is +3 group five is -3 group six is -2 group seven is -1
i would solve the whole thing but its unclear hope this helps tho
2H2O2 decomposes into the products 2H2O + O2(g)
Answer:
The correct answer is A. 140 atm
Explanation:
We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. We convert the unit Celsius into Kelvin:
0 ° C = 273K, 67 ° C = 273 + 67 = 340K; 94 ° C = 273 + 94 = 367K
P1xV1 /T1= P2x V2/T2
P2= ((P1xV1 /T1)xT2)/V2
P2=((88,89atm x 17L/340K)x367K)/12L= <em>135,927625 atm</em>
Answer : The activation energy of the reaction is, 
Solution :
The relation between the rate constant the activation energy is,
![\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7BK_2%7D%7BK_1%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial rate constant = 
= final rate constant = 
= initial temperature = 
= final temperature = 
R = gas constant = 8.314 kJ/moleK
Ea = activation energy
Now put all the given values in the above formula, we get the activation energy.
![\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7B8.75%5Ctimes%2010%5E%7B-3%7DL%2Fmole%5Ctext%7B%20s%7D%7D%7B4.55%5Ctimes%2010%5E%7B-5%7DL%2Fmole%5Ctext%7B%20s%7D%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20%288.314kJ%2FmoleK%29%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7B468K%7D-%5Cfrac%7B1%7D%7B531K%7D%5D)

Therefore, the activation energy of the reaction is, 