Water is the only one of these that would work by process of elimination.
Flame colors are produced from the movement of the electrons in the metal ions present in the compounds. When you heat it, the electrons gain energy and can jump into any of the empty orbitals at higher levels Each of these jumps involves a specific amount of energy being released as light energy, and each corresponds to a particular color. As a result of all these jumps, a spectrum of colored lines will be produced. The color you see will be a combination of all these individual colors.
Answer:
b. chlorine
Explanation:
it has the highest atomic radius
<span>n this order, Ď=1.8gmL, cm=0.5, and mole fraction = 0.9
First, let's start with wt%, which is the symbol for weight percent. 98wt% means that for every 100g of solution, 98g represent sulphuric acid, H2SO4.
We know that 1dm3=1L, so H2SO4's molarity is
C=nV=18.0moles1.0L=18M
In order to determine sulphuric acid solution's density, we need to find its mass; H2SO4's molar mass is 98.0gmol, so
18.0moles1Lâ‹…98.0g1mole=1764g1L
Since we've determined that we have 1764g of H2SO4 in 1L, we'll use the wt% to determine the mass of the solution
98.0wt%=98g.H2SO4100.0g.solution=1764gmasssolution→
masssolution=1764gâ‹…100.0g98g=1800g
Therefore, 1L of 98wt% H2SO4 solution will have a density of
Ď=mV=1800g1.0â‹…103mL=1.8gmL
H2SO4's molality, which is defined as the number of moles of solute divided by the mass in kg of the solvent; assuming the solvent is water, this will turn out to be
cm=nH2SO4masssolvent=18moles(1800â’1764)â‹…10â’3kg=0.5m
Since mole fraction is defined as the number of moles of one substance divided by the total number of moles in the solution, and knowing the water's molar mass is 18gmol, we could determine that
100g.solutionâ‹…98g100gâ‹…1mole98g=1 mole H2SO4
100g.solutionâ‹…(100â’98)g100gâ‹…1mole18g=0.11 moles H2O
So, H2SO4's mole fraction is
molefractionH2SO4=11+0.11=0.9</span>
Answer:
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
Explanation:
K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)
The complete ionic equation for the above equation can be written as follow:
In solution, K2CO3 and CuF will dissociate as follow:
K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)
CuF(aq) —› Ca^2+(aq) + 2F¯(aq)
Thus, we can write the complete ionic equation for the reaction as shown below:
K2CO3(aq) + 2CuF(aq) —›
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
