Answer:
1. Ionic bond
2. High melting point and high boiling point for ionic bonds while covalent bonds have low melting and boiling point.
3. The similarity is that ionic and covalent bonding lead to the creation of stable molecules.
4. 4Fe + 3O2 → 2Fe2O3
5. It uses the process of fission.
6. Fission involves the splitting of radioactive elements into smaller particles/compounds while Fusion involves combining of two or more atomic nuclei to form one or more different atomic nuclei and subatomic particles.
7. Nuclear power plants produce little to no greenhouse gas.
Nuclear power plants produce a large amount of energy for a small mass of fuel.
Nuclear is less expensive.
The answer is letter C.
Because we classify something as a star when it is: a large ball of gas that undergoes nuclear fusion. Given this definition, a comet is not a star. A comet is a ball of ice and dirt hurtling through space, it shines only because it reflects ligh
Because subatomic particles ARE what make up atoms.
Answer:
Conociendo el volumen de solución, masa de soluto y su masa molar, es posible determinar: B) Concentración molar
La molaridad es la relación entre el número de moles de soluto y los litros de solución. Más:
M = No moles de solución de soluto / volumen (L)
Y a su vez los moles de soluto se encuentran por:
No moles de soluto = masa soluto / masa molar soluto
Answer:
![\large \boxed{\text{b. 1.5 h}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7Bb.%201.5%20h%7D%7D)
Explanation:
1. Calculate the rate constant
The integrated rate law for first order decay is
![\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%20%28%5Cdfrac%7BA_%7B0%7D%7D%7BA_%7Bt%7D%7D%5Cright%20%29%20%3D%20kt)
where
A₀ and A_t are the amounts at t = 0 and t
k is the rate constant
![\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cln%20%5Cleft%20%28%5Cdfrac%7B500%7D%7B31%7D%5Cright%29%20%26%20%3D%20%26%20k%20%5Ctimes%206%5C%5C%5C%5C%5Cln%2016.1%20%26%20%3D%20%26%206k%5C%5C2.78%26%20%3D%26%206k%5C%5Ck%20%26%20%3D%20%26%20%5Cdfrac%7B2.78%7D%7B6%7D%5C%5C%5C%5C%26%20%3D%20%26%200.463%20%5Ctext%7B%20h%7D%5E%7B-1%7D%5C%5C%5Cend%7Barray%7D)
2. Calculate the half-life
![t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%3D%20%5Cdfrac%7B%5Cln2%7D%7Bk%7D%20%3D%20%5Cdfrac%7B%5Cln2%7D%7B%5Ctext%7B0.463%20%20h%7D%5E%7B-1%7D%7D%20%3D%20%5Ctextbf%7B1.5%20h%7D%5C%5C%5C%5C%20%5Ctext%7BThe%20half-life%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B1.5%20h%7D%7D%24%7D)