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Svetach [21]
3 years ago
7

When a 5.00-g sample of rbbr is dissolved in water in a calorimeter?

Chemistry
1 answer:
lilavasa [31]3 years ago
7 0
To answer this question, I found a similar problem as shown in the picture below. Let's use the given information: Heat capacity = 3.565 kJ/K, ΔT = 0.26 K. Also, the molar mass of RbBr is 165.372 g/mol.

The solution is as follows:

Molar Heat of Solution = (3.565 kJ/K)(0.26 K)(1/ 5 g)(165.372 g/mol)
<em>Molar Heat of Solution = 30.66 kJ/mol</em>

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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, <em>V</em> is the volume of the gas, <em>T</em> is Kelvin temperature, and <em>k</em> 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 .

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2 years ago
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ObIel WiLll unt COl.. USSMS A certain chemical reaction releases 31.2 kJ/g of heat for each gram of reactant consumed. How can y
aleksandr82 [10.1K]

Answer:

The expression to calculate the mass of the reactant is m = \frac{1.080kJ}{31.2kJ/g}

Explanation:

<em>The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant.</em> To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:

1,080J.\frac{1kJ}{10^{3}J } =1.080kJ

Then,

1.080kJ=31.2\frac{kJ}{g} .m\\m = \frac{1.080kJ}{31.2kJ/g}

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Comare and contrast physical property and chemical property
Nimfa-mama [501]
A physical property is a quality or condition of a substance that can be observed or measured without changing the substance's composition. a chemical property is the ability of a substance to undergo a specific chemical change.
8 0
3 years ago
What are isotopes please help
NikAS [45]

Answer:

Isotopes are atoms of the same element with the same number of protons but different number of neutrons.

Hope this helps.

6 0
3 years ago
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Which location of the moon relative to the sun and earth may produce a warning (getting smaller nightly) half moon
scZoUnD [109]

Answer:

The moon to the direct right of the Earth relative to the Sun

Explanation:

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