<u>Answer:</u> The wavelength of the flame is 462 nm and color of cesium flame is blue.
<u>Explanation:</u>
To calculate the wavelength, we use Planck's equation, which is:

where,
E = Energy of 1 photon = 
h = Planck's constant = 
c = speed of light = 
= wavelength = ?
Putting values in above equation, we get:

The range of wavelength of blue light lies in range of 500 nm - 435 nm
The calculated wavelength lies in the above range. So, the color of the cesium flame is 462 nm
Hence, the wavelength of the flame is 462 nm and color of cesium flame is blue.
Answer:
The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons's law). ... The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle's law).
Respuesta:
21.8 L
Explicación:
Paso 1: Escribir la reacción balanceada
CaCO₃ ⇒ CaO + CO₂
Paso 2: Convertir 450 g de CaCO₃ a moles
La masa molar de CaCO₃ es 100.09 g.
450 g × (1 mol/100.09 g) = 4.50 mol
Paso 3: Calcular los moles de CO₂ que se forman a partir de 4.50 moles de CaCO₃
La relación molar de CaCO₃ a CO₂ es 1:1. Los moles de CO₂ formados son 1/1 × 4.50 mol = 4.50 mol.
Paso 4: Convertir la temperatura a Kelvin
Usaremos la siguiente expresión.
K = °C + 273.15 = 200°C + 273.15 = 473 K
Paso 5: Calcular el volumen de CO₂
Usaremos la ecuación del gas ideal.
P × V = n × R × T
V = n × R × T / P
V = 4.50 mol × (0.082 atm.L/mol.K) × 473 K / 8 atm
V = 21.8 L
Answer:
96%
Explanation:
Step 1: Write the balanced neutralization reaction
Al(OH)₃ + 3 HCl ⇒ AlCl₃ + 3 H₂O
Step 2: Calculate the theoretical yield of AlCl₃
According to the balanced equation, the mass ratio of Al(OH)₃ to AlCl₃ is 81.03:133.34.
28 g Al(OH)₃ × 133.34 g AlCl₃/81.03 g Al(OH)₃ = 46 g AlCl₃
Step 3: Calculate the percent yield of AlCl₃
The real yield of AlCl₃ is 44 g. We can calculate the percent yield using the following expression.
%yield = real yield / theoretical yield × 100%
%yield = 44 g / 46 g × 100% = 96%
Answer is: <span>B. taking photographs.
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