Moles in 72 grams of NaHCO3

Please help. Idk what to do. Please

Inga [223]

Answer:

7. orbit

8. A comet

9.gas giant

10.moons

11.gravity

12.meteoroids

13.craters

14.super sized

Explanation: idk if 14 is right but yea your welcome

5 0

3 years ago

Calculate the frequency of the n=2 line in the lyman series of hydrogen

Alona [7]

Answer:

Approximately $2.47\times 10^{15}\; \rm Hz$ .

Explanation:

The Lyman Series of a hydrogen atom are due to electron transitions from energy levels $n \ge 2$ to the ground state where $n = 1$ . In this case, the electron responsible for the line started at $n = 2$ and transitioned to

A hydrogen atom contains only one electron. As a result, Bohr Model provides a good estimate of that electron's energy at different levels.

In Bohr's Model, the equation for an electron at energy level $n$ (

$\displaystyle - \frac{k\, Z^2}{n^2}$ (note the negative sign in front of the fraction,)

where

$k = 2.179 \times 10^{-18}\; \rm J$ is a constant.

$Z$ is the atomic number of that atom. $Z = 1$ for hydrogen.
$n$ is the energy level of that electron.

The electron that produced the $n = 2$ line was initially at the

$\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}$ .

The electron would then transit to energy level $n = 1$ . Its energy would become:

$\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}$ .

The energy change would be equal to

$\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}$ .

That would be the energy of a photon in that $n = 2$ spectrum line. Planck constant $h$ relates the frequency of a photon to its energy:

$E = h \cdot f$ , where

$E$ is the energy of the photon.
$h \approx 6.62607015\times 10^{-34}\; \rm J \cdot s$ is the Planck constant.
$f$ is the frequency of that photon.

In this case, $E \approx 1.63425 \times 10^{-18}\; \rm J$ . Hence,

$\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}$ .

Note that $1 \; \rm Hz = 1 \; \rm s^{-1}$ .

6 0

3 years ago

What is the simplest formula of a solid containing a, b, and c atoms in a cubic lattice in which the a atoms occupy the corners,

Ilya [14]

Answer:

c. abc₃

Step-by-step explanation:

1. a atoms

There are eight corners, each containing an a atom.

No. of a atoms = 8 × ⅛

No. of a atoms = 1

=====

2. b atoms

There is one central b atom.

=====

3. c atoms

There are six faces, each containing a c atom.

No. of a atoms = 6 × ½

No. of c atoms = 3

=====

4. Formula

The simplest formula is abc₃.

4 0

3 years ago

How many liters of a 0.0550 M NaF solution contain 0.163 moles of NaF?

Kobotan [32]

Answer : The volume of solution will be 2.96 liters.

Explanation :

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

In this question, the solute is NaF.

Now put all the given values in this formula, we get:

$0.0550M=\frac{0.163mole}{\text{Volume of solution (in L)}}$

$\text{Volume of solution (in L)}=\frac{0.163mole}{0.0550M}$

$\text{Volume of solution (in L)}=2.96L$

Therefore, the volume of solution will be 2.96 liters.

3 0

2 years ago

Although Group 7A ions may have different oxidation states depending on which other elements are in the compound, the most commo

frutty [35]

1-. Because Group 7A elements have 7 electrons in their valence shell, they need 1 more electron to complete the octet. Thus, it is generally true that all Group 7 electrons have an oxidation number of 1-. This also explains why electron affinity is largely the highest for Group 7A and lowest for Group 1A.

6 0

3 years ago