According to this formula :
㏑[A] /[Ao] = - Kt
when we have Ao = 0.3 m
and K =0.46 s^-1
t = 20min = 0.2 x 60 =12 s
So by substitution :
㏑[A] / 0.3 = - 0.46 * 12
㏑[A] / 0.3 = - 5.52
by taking e^x for both side of the equation we can get [A]
∴[A] = 0.0012 mol dm^-3
It respresents the higher energy level than 627nm .
<h3>What is a emission line ? </h3>
Emission lines are the glowing hot gas emits lines of light whereas absorption line refers to the tendency of cool atmospheric gas to absorb the same line of light.Some lights produce dark band when the light passes through gas in the atmosphere . There are two line spectrum and absorption.
spectrum is an excitement of electrons from lower to higher energy levels and when it comes back it releases energy in the terms of colourful lights .
It represents the higher energy levels than 627nm because Energy is inversly proportional to wavelength .
to learn more about Emission lines click here
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It has 4 significant figures. If you see the Zero after the one decimal point you don’t count that and instead just started at 6
Answer:
4.549 kg.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 2 x 10⁴ kPa/101.325 = 197.4 atm).
V is the volume of the gas in L (V = 20.0 L).
n is the no. of moles of the gas in mol (n = ??? mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 23° C + 273 = 296 K).
<em>∴ n = PV/RT =</em> (197.4 atm)(20.0 L)/(0.0821 L.atm/mol.K)(296 K) = <em>162.5 mol.</em>
- To find the mass of N₂ in the cylinder, we can use the relation:
<em>mass of N₂ = (no. of moles of N₂)*(molar mass of N₂) = </em>(162.5 mol)*(28.0 g/mol) = <em>4549 g = 4.549 kg.</em>
