Fe3(PO4)2 Iron(II) Phosphate
Fe3(PO3)2 iron(II) Phosphite
Fe(PO4) Iron(III) Phosphate
FePO3 (Iron (iii) phosphite.
Explanation:
The ions given are
Fe^3+
Fe^2+
PO4^3-
PO3^3-
The ionic compounds or electrically neutral compounds that can be formed from the cations and anions given are :
Fe^3+ + PO3^3-= Fe3(PO4)2 the ions are balanced. the compound formed is Iron(II) Phosphate
Fe^2++PO4^3-= Fe3(PO3)2 is iron(II)Phosphite
The compound formed is balanced with the charges provided on the ions.
Fe^3+ + PO4^3-= Fe(PO4) the charge on the anion and cation are equal.
THE COMPOUND FORMED IS Iron(III) Phosphate
Fe^3++ PO3^3= FePO3
Fe^3+ + PO^3-= FePO3 (Iron (iii) phosphite.
Answer: C. 146 g
Explanation:
To calculate the moles, we use the equation:
Thus
For sodium nitrate
moles = 1.72
Molar mass of sodium nitrate = 85g/mol
Putting values in above equation, we get:
Answer:Explanation:The higher the frequency, the faster the oscillations and thus the higher the energy. Therefore, the highest-frequency ultra-violet light (or the lowest wavelength) is violet. However, the highest-frequency visible light would have to be roughly blue.
Answer: (4) C-14 to 6-12
Explanation:
Carbon dating is used to estimate the age of wooden trees which use carbon dioxide. Carbon has three naturally occurring isotopes.C-12 and C-13 are stable but C-14 is radioactive. C-14 decays to nitrogen-14 and has a half-life of 5,730 years.
Wooden trees naturally incorporate both the C-12 and C-14 during their lifetimes. When wooden trees die, it stops consuming more of radioactive carbon while the C-14 which is already present in them continues to decay back into nitrogen. With the help of C-12 to C-14 ratio, we can estimate the age of wooden objects.
Answer:
10.5grams
Explanation:
Molarity = number of moles (n)/ volume (V)
According to this question;
Volume = 750 mL = 750/1000 = 0.75L
Molarity = 0.35M
number of moles (n) = molarity × volume
n = 0.35 × 0.75
n = 0.2625mol
Using mole = mass/molar mass
Molar Mass of NaOH = 23 + 16 + 1
= 40g/mol
mole = mass/molar mass
0.2625 = mass/40
mass = 10.5grams
10.5 grams are needed to prepare 0.75L of a 0.35 M solution of NaOH.