Answer: The difference between an element and a compound is that an element is a pure substance that is made of only one element, but a compound is 2 or more elements together
Explanation:
Answer:
Making oxygen
Oxygen can be made from hydrogen peroxide, which decomposes slowly to form water and oxygen:
hydrogen peroxide → water + oxygen
2H2O2(aq) → 2H2O(l) + O2(g)
The rate of reaction can be increased using a catalyst, manganese(IV) oxide. When manganese(IV) oxide is added to hydrogen peroxide, bubbles of oxygen are given off.
Apparatus arranged to measure the volume of gas in a reaction. Reaction mixture is in a flask and gas travels out through a pipe in the top and down into a trough of water. It then bubbles up through a beehive shelf into an upturned glass jar filled with water. The gas collects at the top of the jar, forcing water out into the trough below.
To make oxygen in the laboratory, hydrogen peroxide is poured into a conical flask containing some manganese(IV) oxide. The gas produced is collected in an upside-down gas jar filled with water. As the oxygen collects in the top of the gas jar, it pushes the water out.
Instead of the gas jar and water bath, a gas syringe could be used to collect the oxygen.
That is a graph read both sides and you see speed and distance which must mean that someone is graphing peoples speed.
Answer:
Explanation:
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In this case, for this stoichiometry-based problem, it is firstly necessary to realize that the decomposition of ammonium dichromate is given by:
Thus, since the mole ratio between ammonium dichromate and the gaseous nitrogen (molar mass = 28.02 g/mol) is 1:1, we can compute the produced mass of the latter via stoichiometry as shown below:
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Answer:
a). P = 688 atm
b). P = 1083.04 atm
c).Δ G = 16.188 J/mol
Explanation:
a). Fugacity 'f' can be calculated from the following equations :
where, P = pressure , Z = compressibility
Now, the virial equation is :
........(1)
Also, PV=ZRT for real gases .......(2)
∴ 
So from the fugacity equation ,
Putting the value of P = 500 atm in the above equation, we get,
f = 688 atm
b). Given f = 2P
∴ P = 1083.04 atm
c). dG = Vdp -S dt at constant temperature, dT = 0
Therefore, dG = V dp
![$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$](https://tex.z-dn.net/?f=%24%5CDelta%20G%3DR%5B%5Cln%5Cfrac%7BP_2%7D%7BP_1%7D%2B6.4%20%5Ctimes%2010%5E%7B-4%7D%28P_2-P_1%29%5D%24)
![$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$](https://tex.z-dn.net/?f=%24%5CDelta%20G%3D8.314%5Ctimes%20298%5B%5Cln%5Cfrac%7B500%7D%7B1%7D%2B6.4%20%5Ctimes%2010%5E%7B-4%7D%28500-1%29%5D%24)
Δ G = 16.188 J/mol
