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3 years ago

6

You have two solutions, both with a concentration of 0.1M. Solution A contains a weak acid with a pKa of 5. The pH of solution A

is 3. Solution B contains a weak acid with a pKa of 9. The pH of solution B is:

1 answer:

viktelen [127]3 years ago

3 0

Answer:guguhi

Explanation:

Guggugj

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Which pH corresponds to the equivalence point for the graph below?

Evgen [1.6K]

Answer 10
Explication nose

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3 years ago

What is the molarity of the solution formed by dissolving 80. G of NAOH(s) into water to give a total volume of 4.00 l

Mrrafil [7]

Answer:

0.5 M

Explanation:

From the question given above, the following data were obtained:

Mass of NaOH = 80 g

Volume of solution = 4 L

Molarity =?

Next, we shall determine the number of mole in 80 g of NaOH. This can be obtained as follow:

Mass of NaOH = 80 g

Molar mass of NaOH = 23 + 16 + 1

= 40 g/mol

Mole of NaOH =?

Mole = mass / molar mass

Mole of NaOH = 80 / 40

Mole of NaOH = 2 moles

Finally, we shall determine the molarity of the solution. This can be obtained as follow:

Mole of NaOH = 2 moles

Volume of solution = 4 L

Molarity =?

Molarity = mole / Volume

Molarity = 2/4

Molarity = 0.5 M

Therefore, the molarity of the solution is 0.5 M.

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3 years ago

Ammonia and sulfuric acid react to form ammonium sulfate. Determine the starting mass of each reactant if 20.3 g of ammonium sul

strojnjashka [21]

Hope you are able to see

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3 years ago

ON a meter stick, one centimeter is generally the___ a)length of the whole stick b)length of the one hundred sticks c)distance b

Elena-2011 [213]

c)distance between two of the numbered lines

Explanation:

On a meter stick, a centimeter is usually the distance between two of the numbered lines. A meter stick is a device for measuring the length of an object.

A hundred centimeters makes up a meter and this is the number of divisions on a meter stick.
Each distance between numbered vertical lines is a centimeter apart.
On a meter stick that is 1m long, there are 100cm.

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lengths brainly.com/question/4687435

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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