Calcium forms an ion with a positive 2 charge and chlorine forms an ion with a negative one charg, so the formula is <span>CaC<span>l2</span></span>
Group 1 metals and group 2 metals form positive ions by losing 1 and 2 electrons respectively. Non-metals in group 17 gain 1, group 16 gain 2 and group 15 gain 3. Elements which lose electrons form positive ions while elements that gain electrons form negative ions.
To write a formula, you must balance charges so the overall charge is zero. A simple way to do this is to swap the # of the ion's charge and make it the subscript of the other ion. However, leave off the number 1 and reduce to lowest whole number ratio.
Answer:
The net energy is 2.196 eV
Explanation:
Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.
Using:
ΔE = h*c*(1/λ
- 1/λ
)
where:
ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602*
J.
h = 4.135*
eVs
c = 3*
m/s
λ
= 300 nm = 300*
m
λ
= 640 nm = 640*
m
Thus:
ΔE = 4.135*
eVs*3*
m/s*(
)
ΔE = 4.135*
*3*
*1.77*
eV = 2.196 eV
Answer:
the mixture or solution is called homogeneous mixture
Answer:
The answer to your question is
1.-Fe₂O₃
2.- 280 g
3.- 330 g
Explanation:
Data
mass of CO = 224 g
mass of Fe₂O₃ = 400 g
mass of Fe = ?
mass of CO₂
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
1.- Calculate the molar mass of Fe₂O₃ and CO
Fe₂O₃ = (56 x 2) + (16 x 3) = 160 g
CO = 12 + 16 = 28 g
2.- Calculate the proportions
theoretical proportion Fe₂O₃ /3CO = 160/84 = 1.90
experimental proportion Fe₂O₃ / CO = 400/224 = 1.78
As the experimental proportion is lower than the theoretical, we conclude that the Fe₂O₃ is the limiting reactant.
3.- 160 g of Fe₂O₃ --------------- 2(56) g of Fe
400 g of Fe₂O₃ --------------- x
x = (400 x 112) / 160
x = 280 g of Fe
4.- 160 g of Fe₂O₃ --------------- 3(44) g of CO₂
400 g of Fe₂O₃ -------------- x
x = (400 x 132)/160
x = 330 gr
Based on Beer-Lambert's Law,
A = εcl ------(1)
where A = absorbance
ε = molar absorptivity
c = concentration
l = path length
Step 1: Calculate the concentration of the diluted Fe3+ standard
Use:
V1M1 = V2M2
M2 = V1M1/V2 = 10 ml*6.35*10⁻⁴M/55 ml = 1.154*10⁻⁴ M
Step 2 : Calculate the concentration of the sample solution
Based on equation (1) we have:
A(Fe3+) = ε(1.154*10⁻⁴)(1)
A(sample) = ε(C)(4.4)
It is given that the absorbances match under the given path length conditions, i.e.
ε(1.154*10⁻⁴)(1) = ε(C)(4.4)
C = 0.262*10⁻⁴ M
This is the concentration of Fe3+ in 100 ml of well water sample
Step 3: Calculate the concentration of Fe3+ in the original sample
Use V1M1 = V2M2
M1 = V2M2/V1 = 100 ml * 0.262*10⁻⁴ M/35 ml = 7.49*10⁻⁵M
Ans: Concentration of F3+ in the well water sample is 7.49*10⁻⁵M