I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y,
for the following reason:
If you look at the combining masses of X and Y in
each of the two compounds,
The first compound contains 0.25g of X combined with
0.75g of Y
so the ratio (by mass) of X to Y = 1 : 3
The second compound contains 0.33 g of X combined with
0.67 g of Y
so the ratio (by mass) of X to Y = 1 : 2
Now, you suppose to prepare each of these two
compounds, starting with the same fixed mass of element Y ( I will choose 12g
of Y for an easy calculation!)
The first compound will then contain 4g of X and 12g
of Y
The second compound will then contain 6g of X and
12g of Y
<span>The ratio which combined
the masses of X and the fixed mass (12g) of Y
= 4 : 6
<span>or 2 : 3 </span>
So, the ratio of MOLES of X which combined with the
fixed amount of Y in the two compounds is also = 2 : 3 </span>
The two compounds given with the plausible formula must therefore contain
the same ratio.
Answer:
(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔2Na[Al(OH)4](aq) + 3H2(g)
∴ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11
(2) H2O(l) + SO3(g) ↔ H2SO4(aq)
∴ Kc = [ H2SO4 ] / PSO3 = 0.0123
(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)
∴ Kc = Kc = 1 / PO2∧6
Explanation:
(1) 2Al(s) + 2NaOH(aq) + 6H2O(l) ↔ 2Na[Al(OH)4](aq) + 3H2(g)
∴ O / Al: 0 → +2 ≡ 2e-
Na: +1 → +2
∴ R / H: +1 → 0
2 - Al - 2
2 - Na - 1
8 - O - 8
14 - H - 14
⇒ Kc = ( PH2³ * [Na[Al(OH)4]² ) / [NaOH]² = 11
(2) H2O(l) + SO3(g) ↔ H2SO4(aq)
1 - S - 1
4 - O - 4
2 - H - 2
⇒ Kc = [ H2SO4 ] / PSO3 = 0.0123
(3) 2P4(s) + 6O2(g) ↔ 2P4O6(s)
8 - P - 8
12 - O - 12
⇒ Kc = 1 / PO2∧6
Answer:
photosyntheisis
Explanation:
The process by which plants and some other organisms capture enrgy in sunlight and use it to make food (Btw ik this is right becuase i did the same thing and i have the definition of what it is and other definitions!
Answer:
hexaaquairon(III) trinitrate
Answer:
A. Generally ionic compounds are formed metal and non metal elements since they forms opposite charged ions due to difference in their Electronegativity values.
NaCl, MgCl2, Ca(OH)2 etc are ionic compounds.
B. Generally transition elements (D elements) forms cations with different charges or simply they exist in different oxidation states due to availability of empty d orbitals.
Examples are Iron exsits as Fe+2 and Fe+3
Mn exists as Mn+2 , Mn+4, Mn+6 ,Mn+7 etc.
