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adelina 88 [10]
3 years ago
11

What are the elements of NaCl

Physics
1 answer:
Lana71 [14]3 years ago
4 0

The elements in NaCl are Na -sodium and Cl - chlorine

<u>Explanation:</u>

  • sodium chloride (NaCl) is also known as common salt. The molar ratio of NaCl is 1:1 and has the molar mass of 22.99 and 35.45 respectively. The molar mass of NaCl is 58.443 g/mol. It is a odorless and colourless cubic crystal.
  • It has the melting point of 800.7°C and has the boiling point of 1465°C. It has a face centered cubic crystal structure.The elements in NaCl are Na -sodium and Cl - chlorine

                                                        2Na+Cl_{2} → 2NaCl

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A piston of volume 0.1 m3 contains two moles of a monatomic ideal gas at 300K. If it undergoes an isothermal process and expands
seropon [69]

Answer:

the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)

Explanation:

assuming ideal gas behaviour of the gas , the equation for ideal gas is

P*V=n*R*T

where

P = absolute pressure

V= volume

T= absolute temperature

n= number of moles of gas

R= ideal gas constant = 8.314 J/mol K

P=n*R*T/V

the work that is done by the gas is calculated through

W=∫pdV=  ∫ (n*R*T/V) dV

for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:

W=∫pdV=  ∫ (n*R*T/V) dV =  n*R*T  ∫(1/V) dV = n*R*T * ln (V₂/V₁)

since

P₁=n*R*T/V₁

P₂=n*R*T/V₂

dividing both equations

V₂/V₁ = P₁/P₂

W= n*R*T * ln (V₂/V₁)  = n*R*T * ln (P₁/P₂ )

replacing values

P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa

since P₂ = 1 atm = 101325 Pa

W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J

5 0
3 years ago
The operating temperature of a tungsten filament in an incandescent light bulb is 2450 K, and its emissivity is 0.350. Find the
natulia [17]

Answer:

The value is   A =  2.80 *10^{-4} \  m^2

Explanation:

From the question we are told that

The  operating temperature is  T  =  2450 \  K

The emissivity is  e =  0.350

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Where  \sigma is the Stefan Boltzmann constant  with value  

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So

     A =  \frac{200}{0.350 *  5.67*10^{-8} *  2450^{4}}

     A =  2.80 *10^{-4} \  m^2

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