The different types of electromagnetic radiation shown in the electromagnetic spectrum consists of radio waves, microwaves, infrared waves, visible light, ultraviolet radiation, X-rays, and gamma rays.
Pete because 2.5 x 2.3 is faster
Answer:
The answer to your question is: Volume = 29.79 cm³
Explanation:
Data
Density of gold = 19.3 g/cm3
Mass = 575 grams
Formula
density = mass / volume
Volume = mass/density
Volume = 575 g / 19.3 g/cm3
Volume = 29.79 cm³
Answer : The value of 
is 28.97 kJ/mol
Explanation :
To calculate of the reaction, we use clausius claypron equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= vapor pressure at temperature 
= 462.7 mmHg
= vapor pressure at temperature 
= 140.5 mmHg
= Enthalpy of vaporization = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = ![-21.0^oC=[-21.0+273]K=252K](https://tex.z-dn.net/?f=-21.0%5EoC%3D%5B-21.0%2B273%5DK%3D252K)
= final temperature = ![45^oC=[-41.0+273]K=232K](https://tex.z-dn.net/?f=45%5EoC%3D%5B-41.0%2B273%5DK%3D232K)
Putting values in above equation, we get:
![\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B140.5mmHg%7D%7B462.7mmHg%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B252%7D-%5Cfrac%7B1%7D%7B232%7D%5D%5C%5C%5C%5C%5CDelta%20H_%7Bvap%7D%3D28966.6J%2Fmol%3D28.97kJ%2Fmol)
Therefore, the value of is 28.97 kJ/mol
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
