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3 years ago

Besides water, what is the product of a Neutalization Reaction between HBr and CsOH?

Chemistry

2 answers:

Luba_88 [7]3 years ago

5 0

I don’t know dawg but I’m just answering this so I can complete the steps good luck tho

BlackZzzverrR [31]3 years ago

3 0

Answer:

The answer to your question is the letter d. CsBr

Explanation:

Data

HBr reacts with CsOH

A neutralization reaction occurs between an acid and a base and the products will always be water and a binary salt.

In this reaction, the acid is HBr and the base is CsOH, then the products will be water and Cesium bromide.

Balanced chemical reaction

HBr + CsOH ⇒ CsBr + H₂O

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What is the wavelength of this radiation in centimeters?

Allushta [10]

multiply by 100.4.36×10-5cm

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3 years ago

In trying to control fall armyworms in crops, an Agriculture extension officer applied cypermethrin which was prepared by dissol

sveta [45]

Answer:

Mole fraction for C₂₂H₁₉Cl₂NO₃ = 0.0086

Explanation:

Mole fraction remains a sort of concentration. It indicates:

moles of solute / (moles of solute + moles of solvent)

Moles of solute / Total moles.

Solute: Cypermethrin → C₂₂H₁₉Cl₂NO₃

Solvent: Water (PM = 18g/mol)

We calculate moles from solvent: 1000g /18 g/mol = 55.5 moles

We calculate PM for C₂₂H₁₉Cl₂NO₃

12g/mol . 22 + 1g/mol . 19 + 35.45 g/mol . 2+ 14g/mol + 16g/mol . 3 = 416 g/m

Moles of solute: 200 g / 416g/mol = 0.481 moles

Total moles: 0.481 + 55.5 = 55.98 moles

Mole fraction for C₂₂H₁₉Cl₂NO₃ = 0.481 moles / 55.98 moles = 0.0086

8 0

3 years ago

In an aqueous solution of a certain acid the acid is 4.4% dissociated and the pH is 3.03. Calculate the acid dissociation consta

NeX [460]

Answer:

4.1x10⁻⁵

Explanation:

The dissociation of an acid is a reversible reaction, and, because of that, it has an equilibrium constant, Ka. For a generic acid (HA), the dissociation happens by:

HA ⇄ H⁺ + A⁻

So, if x moles of the acid dissociates, x moles of H⁺ and x moles of A⁻ is formed. the percent of dissociation of the acid is:

% = (dissociated/total)*100%

4.4% = (x/[HA])*100%

But x = [A⁻], so:

[A⁻]/[HA] = 0.044

The pH of the acid can be calcualted by the Handersson-Halsebach equation:

pH = pKa + log[A⁻]/[HA]

3.03 = pKa + log 0.044

pKa = 3.03 - log 0.044

pKa = 4.39

pKa = -logKa

logKa = -pKa

Ka = $10^{-pKa}$

Ka = $10^{-4.39}$

Ka = 4.1x10⁻⁵

4 0

3 years ago

Is the process represented below an example of a physical or a chemical

Basile [38]

It is a physical change because only the states as being changes, not the actual bonds in the compound.

6 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

4 years ago