Answer:
d = 0.93 g/cm³
Explanation:
Given data:
Mass of object = 28 g
Volume of object = 3cm×2cm×5cm
density of object = ?
Solution:
Volume of object = 3cm × 2cm ×5cm
Volume of object = 30 cm³
Density of object:
d = m/v
by putting values,
d = 28 g/ 30 cm³
d = 0.93 g/cm³
When naming an ionic compound, write the name of the cation, which is the metal first. Then, write the name of the anion, which is the nonmetal. However, you remove the last 2-3 letters and replace suffixes.
1. RbF --> Rubidium Fluoride
Change fluorine to fluoride
2. CuO --> Copper (II) Oxide
Change oxygen to oxide. Oxide has a charge of -2. Since no subscripts are written, it means they have the same opposite charge. So, we use Copper (II).
<span>3. (NH</span>₄<span>)</span>₂<span>C</span>₂<span>O</span>₄ ---> Ammonium Oxalate
NH₄ is ammonia, but we change it to ammonium for polyatomic ions.
Answer: 
Explanation:
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= 
= activation energy for the reaction = 262 kJ/mol = 262000J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature =
= final temperature = 
Now put all the given values in this formula, we get
![\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B6.1%5Ctimes%2010%5E%7B-8%7D%7D%7BK_2%7D%29%3D%5Cfrac%7B262000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B775.0K%7D%5D)
Therefore, the value of the rate constant at 775.0 K is
Answer:
the force between charged particles increases when they are in solid form
The arrow doesn’t “point” to anything really, it means that the reactants form (or change) into the products.
