<u>Answer:</u> The wavelength of light is 
<u>Explanation:</u>
To calculate the wavelength of light, we use Rydberg's Equation:
Where,
= Wavelength of radiation
= Rydberg's Constant = 
= Final energy level = 3
= Initial energy level = 6
Putting the values in above equation, we get:
Hence, the wavelength of light is
2.10098*10^47 atoms
Because no. Of atoms = no. Of moles * avogadros no
<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
At the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
So Kc = [H2] [Br2] / [HBr]^2
7.04X10^-2 = [H2][Br] / [HBr]^2
at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root:
∴ √(1/7.04X10^-2)= [HBr] / [H2]^1/2*[Br]^1/2
∴ Kc for the second reaction = √(1/7.04X10^-2) = 3.769
<span>(A)hydrochloric acid + silver nitrate
HCl(aq) + AgNO3(aq) -----> AgCl(s) +HNO3(aq)
</span><span>(B)hydrochloric acid + sodium hydroxide
</span><span>HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)
</span><span>(C)calcium chloride + silver nitrate
CaCl2(aq) + AgNO3(aq) ----> </span>AgCl(s) +Ca(NO3)2(aq)
<span>(D)sodium chloride + silver nitrate
</span>NaCl(aq) + AgNO3(aq) ----> AgCl(s) +NaNO32(aq)
AgCl is a white precipitate.
In (B) no precipitate was formed, so answer is B.
