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3 years ago

What is the mass of 5.8 moles of H2S

Chemistry

1 answer:

Phoenix [80]3 years ago

6 0

Answer:

Mass = 197.78 g

Explanation:

Given data:

Number of moles = 5.8 mol

Mass of H₂S = ?

Solution:

Formula:

Mass = number of moles × molar mass

molar mass of H₂S = 34.1 g/mol

by putting values,

Mass = 5.8 mol × 34.1 g/mol

Mass = 197.78 g

Thus, 5.8 moles of H₂S have 197.78 g mass.

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What is the mass number of an ion with 107 electrons, 158 neutrons, and a 1 charge?

vredina [299]

No of protons=107+1=108
No. Of neutrons=158
Mass no. =166

6 0

4 years ago

How many grams of NO can be produced if 204 g of NO2 is mixed with 58.1 g of H2O?

Goshia [24]

Answer:

44.4 grams of NO can be produced

Explanation:

Step 1: Data given

Mass of NO2 = 204 grams

Molar mass NO2 = 46.0 g/mol

Mass of H2O = 58.1 grams

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

3NO2 + H2O→ 2HNO3 + NO

Step 3: Calculate moles NO2

Moles NO2 = 204 grams / 46.0 g/mol

Moles NO2 = 4.43 moles

Step 4: Calculate moles H2O

Moles H2O = 58.1 grams / 18.02 g/mol

Moles H2O = 3.22 moles

Step 5: Calculate limiting reactant

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

NO2 is the limiting reactant. It will completely be consumed (4.43 moles). H2O is in excess. there will react 4.43 /3 = 1.48 moles. There will remain 3.22 - 1.48 = 1.74 moles

Step 6: Calculate moles NO

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

For 4.43 moles NO2 we'll have 4.43/3 = 1.48 moles NO

Step 7: Calculate mass NO

Mass NO = 1.48 moles * 30.01 g/mol

Mass NO = 44.4 grams

44.4 grams of NO can be produced

3 0

3 years ago

Which of the following will have the slowest rate of diffusion at a given temperature?

marysya [2.9K]

<h3>Answer:</h3>

Chlorine gas (Cl₂)

<h3>Explanation:</h3>

According to the Graham's law of diffusion, the diffusion rate of a gas is inversely proportional to the square root of its density or molar mass.
Therefore, a lighter gas will diffuse faster at a given temperature compared to a heavy gas.
Consequently, the heavier a gas is then the denser it is and the slower it diffuses at a given temperature and vice versa.

In this case we are given gases, CI₂

, H₂,He and Ne.

We are required to identify the gas that will diffuse at the slowest rate.
In other words we are required to determine the heaviest gas.

Looking at the molar mass of the gases given;

Cl₂- 70.91 g/mol

H₂- 2.02 g/mol

He - 4.00 g/mol

Ne- 20.18 g/mol

Therefore, chlorine gas is the heaviest and thus will diffuse at the slowest rate among the choices given.

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

4 years ago

Can someone plss help me asap !!

ludmilkaskok [199]

I think the answer is barium

6 0

3 years ago