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3 years ago

PLZ HELP THERE IS A FILE ATTACHED WILL GIVE U BRAINLIEST IF U HELP

Chemistry

2 answers:

erastovalidia [21]3 years ago

6 0

Answer:

I think C (Might not be true)

Explanation:

Svetllana [295]3 years ago

5 0

Answer:

Explanation:

Should be D as it follows the pattern on the period table in group 1.

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How much heat is required to raise the temperature of 65.8 grams of water from 31.5ºC to 46.9ºC?

vodomira [7]

Answer: 1,013.32 cal × 4.18 J/cal = 4,235.68 J

Explanation:

1) Data:

Water ⇒ C = 1 cal/g°C

m = 65.8 g

Ti = 31.5°C

Tf = 36.9°C

Heat, Q = ?

2) Formula:

Q = mCΔT

3) Calculations:

Q = 65.8g × 1 cal/g°C × (46.9°C - 31.5°C) = 1,013.2 cal

4) You can convert from calories to Joules using the conversion factor:

1 cal = 4.18 J

⇒ 1,013.32 cal × 4.18 J/cal = 4,235.68 J

3 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

A piece of an unknown substance weighing 124.0 grams is heated in boiling water to 100.0oc. when the substance is placed in a ca

MakcuM [25]

The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J

The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:

7560J= 124g * (100-26)* specific heat

specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C

3 0

3 years ago

Two colleagues are measuring the melting point of a substance. They observe two ranges of melting point, at 134-137 °C and 132-1

marissa [1.9K]

Explanation:

impurities affect both melting point and boiling point

in the case of melting point it lowers the melting point

so I guess one of the samples Ie contained impurities

6 0

3 years ago

How do acquired characteristics differ from adaptations

SOVA2 [1]

Answer:

acquired traits are traits that organisms develops over a lifetime. adaptation are trait that enhances an organism's ability to survive and reproduce in its environment.

5 0

2 years ago