A is often written as an if/then statement. O prediction O law O theory O guess

For an experiment, a student needs a compound that has the following properties.

Neporo4naja [7]

The correct answer is Potassium Chloride.

7 0

2 years ago

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Part A:

uranmaximum [27]

Answer:

Part A = The mass of sulfur is 6.228 grams

Part B = The mass of 1 silver atom is 1.79 * 10^-22 grams

Explanation:

Part A

Step 1: Data given

A mixture of carbon and sulfur has a mass of 9.0 g

Mass of the product = 27.1 grams

X = mass carbon

Y = mass sulfur

x + y = 9.0 grams

x = 9.0 - y

x(molar mass CO2/atomic mass C) + y(molar mass SO2/atomic mass S) = 22.6

(9 - y)*(44.01/12.01) + y(64.07/32.07)

(9-y)(3.664) + y(1.998)

32.976 - 3.664y + 1.998y = 22.6

-1.666y = -10.376

y = 6.228 = mass sulfur

x = 9.0 - 6.228 = 2.772 grams = mass C

The mass of sulfur is 6.228 grams

Part B

Calculate the mass, in grams, of a single silver atom (mAg = 107.87 amu ).

Calculate moles of 1 silver atom

Moles = 1/ 6.022*10^23

Moles = 1.66*10^-24 moles

Mass = moles * molar mass

Mass = 1.66*10 ^-24 moles *107.87

Mass = 1.79 * 10^-22 grams

The mass of 1 silver atom is 1.79 * 10^-22 grams

5 0

2 years ago

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Cell and chromosome characteristics?

rusak2 [61]

Answer:

hope the inserted image will help :)

Explanation:

5 0

3 years ago

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Choose the more reactive metal Te or O

kvasek [131]

Te is the answer between the two

8 0

3 years ago

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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