Answer:
The answer to your question is 7160 cm
Explanation:
Data
diameter = 1 mm
length = ?
amount of gold = 1 mol
density = 17 g/cm³
Process
1.- Get the atomic mass of gold
Atomic mass = 197 g
then, 197g ------------ 1 mol
2.- Calculate the volume of this wire
density = mass/volume
volume = mass/density
volume = 197/17
volume = 5.7 cm³
3.- Calculate the length of the wire
Volume = πr²h
solve for h
h = volume /πr²
radius = 0.05 cm
substitution
h = 5.7/(3.14 x 0.05²)
h = 5.7 / 0.0025
h = 7159.2 cm ≈ 7160 cm
Answer:
The fraction of energy used to increase the internal energy of the gas is 0.715
Explanation:
Step 1: Data given
Cv for nitrogen gas = 20.8 J/K*mol
Cp for nitrogen gas = 29.1 J/K*mol
Step 2:
At a constant volume, all the heat will increase the internal energy of the gas.
At constant pressure, the gas expands and does work., if the volume changes.
Cp= Cv + R
⇒The value needed to change the internal energy is shown by Cv
⇒The work is given by Cp
To find what fraction of the energy is used to increase the internal energy of the gas, we have to calculate the value of Cv/Cp
Cv/Cp = 20.8 J/K*mol / 29.1 J/K*mol
Cv/Cp = 0.715
The fraction of energy used to increase the internal energy of the gas is 0.715
Answer : HazCom
Explanation : Hazard communication which is also known as HazCom, is a set of processes and procedures that every employers and importers must implement in their workplace to effectively communicate hazards associated with chemicals during handling, shipping, and any form of exposure.
The OSHA Hazard Communication Standard is a U.S. regulation which governs the evaluation and communication of hazards associated with chemicals at the workplace. It is typically not attached to any specific chemical container but is stored in the workplace.
25 g of NH₃ will produce 47.8 g of (NH₄)₂S
<u>Explanation:</u>
2 NH₃ + H₂S ----> (NH₄)₂S
Molecular weight of NH₃ = 17 g/mol
Molecular weight of (NH₄)₂S = 68 g/mol
According to the balanced reaction:
2 X 17 g of NH₃ produces 68 g of (NH₄)₂S
1 g of NH₃ will produce 
g of (NH₄)₂S
25g of NH₃ will produce 
of (NH₄)₂S
= 47.8 g of (NH₄)₂S
Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S
