What elements are in h2o

In the reaction of aluminum bromide with ionized sodium bromide which compoiund is the lewis acid

murzikaleks [220]

The answer is A) Aluminum Bromide
hope this helps :)

3 0

3 years ago

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What distinguishes physical properties from chemical properties

ale4655 [162]

A physical property is an aspect of matter that can be observed or measured without changing it. A chemical property may only be observed by changing the chemical identity of a substance.

6 0

3 years ago

A 5.024 mg sample of an unknown organic molecule containing carbon, hydrogen, and nitrogen only was burned and yielded 13.90 mg

Dafna1 [17]

Answer:

C8H17N

Explanation:

Mass of the unknown compound = 5.024 mg

Mass of CO2 = 13.90 mg

Mass of H2O = 6.048 mg

Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:

For carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 13.90 = 3.791 mg

For hydrogen, H:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 6.048 = 0.672 mg

For nitrogen, N:

Mass N = mass of unknown – (mass of C + mass of H)

Mass of N = 5.024 – (3.791 + 0.672)

Mass of N = 0.561 mg

Now, we can obtain the empirical formula for the compound as follow:

C = 3.791 mg

H = 0.672 mg

N = 0.561 mg

Divide each by their molar mass

C = 3.791 / 12 = 0.316

H = 0.672 / 1 = 0.672

N = 0.561 / 14 = 0.040

Divide by the smallest

C = 0.316 / 0.04 = 8

H = 0.672 / 0.04 = 17

N = 0.040 / 0.04 = 1

Therefore, the empirical formula for the compound is C8H17N

8 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

Electrolysis of molten lead bromide is carried out.

bearhunter [10]

Answer:

Read Below

Explanation:

Electrolysis is not possible with solid lead (II) bromide. This is because the ions are held in a three-dimensional lattice, unable to move freely to the electrodes. Melting enables the ions to become mobile and to travel to the respective electrodes.

The bulb won't glow when the electrodes are embedded in solid lead bromide. The bulb will glow when the material surrounding the electrodes is molten lead bromide. When an ionic compound is in the molten (liquid) form the positive and negative ions are free to move around.

Hopes this Helps :D

Brainiest Please

5 0

2 years ago