There aren't any options given but I would say the answer is "continuous spectrum"
Alkaline metal-group 1
Alkaline earth metals-group 2
halogens=group 7
noble gases-group 8 or o
transition metals are the actinides and the lactanides below the periodic table
<span>When one talks about ppm in a liquid solution someone means mg/L so we would not be using the density. This usually means ug/g or mg/kg
0.115 g Na^+ * 10^6 ug/1 g = 115000 ug/g
4.55 L * 1000 mL/1L = 4550 mL
Concentration of Na^+ in ppm:
115000 ug/g /4550 mL = 25.27 pm of sodium ion</span>
Answer:
113 g NaCl
Explanation:
The Ideal Gas Law equation is:
PV = nRT
In this equation,
> P = pressure (atm)
> V = volume (L)
> n = number of moles
> R = 8.314 (constant)
> T = temperature (K)
The given values all have to due with the conditions fo F₂. You have been given values for all of the variables but moles F₂. Therefore, to find moles F₂, plug each of the values into the Ideal Gas Law equation and simplify.
(1.50 atm)(15.0 L) = n(8.314)(280. K)
2250 = n(2327.92)
0.967 moles F₂ = n
Using the Ideal Gas Law, we determined that the moles of F₂ is 0.967 moles. Now, to find the mass of NaCl that can react with F₂, you need to (1) convert moles F₂ to moles NaCl (via the mole-to-mole ratio using the reaction coefficients) and then (2) convert moles NaCl to grams NaCl (via molar mass from periodic table). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator).
1 F₂ + 2 NaCl ---> Cl₂ + 2NaF
Molar Mass (NaCl): 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl): 58.44 g/mol
0.967 moles F₂ 2 moles NaCl 58.44 g
---------------------- x ----------------------- x ----------------------- = 113 g NaCl
1 mole F₂ 1 mole NaCl
