A car moved 20km East and 70km West. What is the distance?

0.446 g of hydrogen gas fills a 5.0 L bag determine the density of hydrogen

Oduvanchick [21]

The density of hydrogen : ρ = 0.0892 g/L

<h3>Further explanation</h3>

Given

mass of Hydrogen : 0.446 g

Volume = 5 L

Required

The density

Solution

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

The unit of density can be expressed in g/cm³, kg/m³, or g/L

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

Input the value :

ρ = m : V

ρ = 0.446 g : 5 L

ρ = 0.0892 g/L

5 0

3 years ago

A sample of neon has a volume of 40.81 m3 at 23.5C. At what temperature, in Kelvins, would the gas occupy 50.00 cubic meters? As

mezya [45]

At $\fbox{\begin \\363 K \end{minispace}}$ temperature, a sample of neon gas will occupy $50.00 \text{ m}^{3}$ volume.

Further Explanation:

The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”

Mathematically the law can be expressed as,

$\fbox{ \begin \\ V \propto T \end{minispace}}$

Or,

$\frac{V}{T}=k$

Here, V is the volume of the gas, T is Kelvin temperature, and k is proportionality constant.

Given information:

The initial volume of neon gas is $40.81 \text{ m}^{3}$ .

The final volume of neon gas is $50.00 \text{ m}^{3}$ .

The initial temperature value is $23.5 \text{ } ^{\circ} \text{C}$ .

To calculate:

The final temperature

Given Condition:

The pressure is constant.

Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

Here,

$V_{1}$ is the initial volume of the gas.

$V_{2}$ is the final volume of the gas.

$T_{1}$ is the initial temperature of the gas.

$T_{2}$ is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

$\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}$ …… (2)

Step 3: Convert the given temperature from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

$T(\text{K}) = T(^{\circ}\text{C}) + 273.15$ …… (3)

Substitute $23.5\text{ }^{\circ} \text{C}$ for $T(^{\circ}\text{C})$ in equation (3) to convert temperature from degree Celsius to Kelvin.

$T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}$

Step 4: Substitute $40.81 \text{ m}^{3}$ for $V_{1}$ , $50.00 \text{ m}^{3}$ for $V_{2}$ and $296.65 \text{ K}$ for $T_{1}$ in equation (2) and calculate the value of $T_{2}$ .

$T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}$

Important note:

The temperature must be in Kelvin.

The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

5 0

3 years ago

Read 2 more answers

Explain why current flowing indicates the presence of ions

seraphim [82]

Explanation:

An electron current, the flow of electrons, contributes to an electric current since the electron 'carries' negative electric charge. ... The flow of ions (either positively or negatively charged) also contributes to an electric current in, for example, the electrolyte of an electrochemical cell.

hope it will work well?!

3 0

3 years ago

PLEASE HELP ME Which of the following is a Brønsted-Lowry base but not an Arrhenius base?

never [62]

Ur answer is going to be NH 3 or D AND PLEASE MARK ME AS BRAINLIEST CAUSE IVE NEVER GOT BRAINLIEST BEFORE AND HAVE A GOOD DAY !

4 0

3 years ago

Match each statement with a layer of the atmosphere. There will be two statements for each layer.

LuckyWell [14K]

The troposphere:

H. This layer can have thunderstorms or clear, sunny skies.

A. The biosphere interacts most with this layer.

The stratosphere:

B. It is the second layer from Earth's surface.

G. Winds are strong and steady in this layer.

The mesosphere:

E. It is heated by the ozone layer beneath it.

D. This layer is where most meteor showers occur.

The thermosphere :

F. It contains the ionosphere and exosphere.

C. It contains layers of single, unmixed gas.

Explanation:

Depending on the Earth's temperature the atmosphere can be separated into layers. The troposphere, the stratosphere, the mesosphere and the thermosphere are those layers. The lowest layer is named as Troposphere (0-10 km from the Earth outer surface), it comprises about 75% of the atmosphere's total air and nearly most the water vapor.

Stratosphere (10-30) includes much of the surface ozone. The change in height temperature arises as this ozone absorbs ultraviolet (UV) radiation from the sun. The temperature in Mesosphere (30-50 Km) declines again with height, hitting a minimum of about -90 ° C at the "mesopause." Above this thermosphere (50-400 Km) is settled which is a area where temperatures rise with height once again. The penetration of intense UV and X-ray radiation from the sun induces this temperature rise.

6 0

3 years ago