Answer:
<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em> net ionic equation: </em>3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Explanation:
The balanced equation is
3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)
<em>Ionic equations: </em>Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
. Indicate the correct formula and charge of each ion. Indicate the correct number of each ion
. Write (aq) after each ion
.Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation
3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em>Net ionic equations: </em>Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.
3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Answer : (C) Hafnium is the most likely identity of the given substance.
Solution : Given,
Mass of given substance (m) = 46.9 g
Volume of given substance (V) = 3.5 
First, find the Density of given substance.
Formula used :

Now,put all the values in this formula, we get
= 13.4 g/
So, we conclude that the density of given substance (13.4 g/
) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/
respectively).
According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.
Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.
So, Hafnium is the most likely element which is the identity of the given substance.
Answer:
See explanation
Explanation:
If we look at the electron configuration closely, we will discover that the element must have had a ground state electron configuration of 2,4.
This is because, the innermost shell usually holds two electrons while the outer shells hold eight electrons each. The four electrons must be accommodated in the second shell in the ground state configuration of the compound.
However, when the atom is excited, one electron from this shell may move to the third shell to give the excited state configuration 2-3-1 as shown in the question.
Answer:
a) 72 °F= 22.22 °C
b) 213.8 °C= 416.84°F
c) 180 °C= 453.15 °K
d) 315 °K= 107.33 °F
e) 1750 °F= 1227.594 °K
f) 0 °K= -459.67 °F
Explanation:
Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:
- Fahrenheit a Celsius:

- Celsius a Fahrenheit: °F= °C*1.8 + 32
- Celsius a Kelvin: °K= °C + 273.15
- Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
- Fahrenheit a Kelvin:

Entonces se obtiene:
a) 72 °F=
=22.22 °C
b) 213.8 °C= 213.8*1.8 + 32= 416.84°F
c) 180 °C= 180°C + 273.15= 453.15 °K
d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F
e) 1750 °F=
= 1227.594 °K
f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F