bromine
Explanation:
halogens are a group of elemnts simlar to eachother
flourine, chlorine, and bromine
Answer:
The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g
Explanation:
Step 1: Given data
The supercritical CO2 has a density of 0.469 g/cm³ (or 0.469 g/mL)
The sample hasa volume of 25.0 mL
Step 2: Calculating mass of the sample
The density is the mass per amount of volume
0.469g/cm³ = 0.469g/ml
The mass for a sample of 25.0 mL = 0.469g/mL * 25.0 mL = 11.725g ≈ 11.7g
The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g
Answer: 12.92g of CoSO4
Explanation:
Molar Mass of CoSO4 = 59 + 32 + (16x4) = 59 + 32 +64 = 155g/mol
Molarity of CoSO4 = 0.303mol/L
Mass conc. In g/L = Molarity x molar Mass
= 0.303x155 = 46.965g/L
275 grams of water = 0.275L of water
46.965g of CoSO4 dissolves in 1L
Therefore Xg of CoSO4 will dissolve in 0.275L i.e
Xg of CoSO4 = 46.965x0.275 = 12.92g
Therefore 12.92g of CoSO4 is needed
It makes it more accessible to everybody
Answer:
Explanation:
What we need to do here is to determine the ratios by using the Rydberg equation starting with the transition to n1 = 1, 2,3, etc and see which one fits the data. Remember the question states that they are series and the wavelengths will be for increasing energy levels.
1/λ = Rh x ( 1/n₁² - 1/n₂²)
Lyman series ( n₁=1 and n₂= 2,3 etc) for the first two lines, the ratios will be:
1/λ₁ /1/λ₂ =(1/1 -1/ 2²) / (1/1 -1/ 3²) ⇒ 0.84 ≠ 0.74 (the first ratio)
For Balmer series n₁ = 2 and n₂ = 3,4,5, etc
1/λ₁ /1/λ₂ =(1/4 -1/3²) / (1/4 -1/4²) ⇒ 0.741 = 0.741 (match!)
Lets use the third line to check our answer:
1/λ₁ /1/λ₂ =(1/4 -1/3²) / (1/4 -1/5²) = 0.66
