Answer:
a) the element belongs to group 2
b) see image attached
c) Magnesium
d) X3N2
Explanation:
Any ion with the formula X^2+ was derived from group 2 elements. The group two element found in period 3 is magnesium alone.
Since the nitride ion has a valency of -3 and X^2+ has a valency of +2 , then a compound formed between X^2+ and the nitride ion will have the formula X3N2.
B.) 2:1 as 32 is the double of single oxygen mass & carbon is of single
Answer:
The mass of a sample of iron that has had 300 J applied to it and heats up from 20 degrees Celsius to 40 degrees Celsius is 32.61 grams.
Explanation:
Calorimetry is the measurement and calculation of the measurement of heat changes exchanged by a body or a system produced in physical and chemical processes.
The sensible heat of a body is the amount of heat received or transferred by a body to produce a change in temperature but without a change in physical state.
The sensible heat in a constant pressure is calculated by:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c, and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial)
In this case:
- c= 0.46

- m= ?
- Tfinal= 40 C
- Tinitial= 20 C
Replacing:
300 J= 0.46
* m* (40 - 20) C
Solving:
300 J= 0.46
* m* 20 C

m= 32.61 g
I'm not sure what type of fuel you're talking about. If you mean fossil fuel then kerosene has the highest density of all of the things you listed.
Answer: The enthalpy change is 34.3 kJ
Explanation:
The conversions involved in this process are :

Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change = ?
m = mass of water = 72.0 g
= specific heat of ice = 
= specific heat of liquid water = 
n = number of moles of water = 
= enthalpy change for fusion = 6010 J/mole
Now put all the given values in the above expression, we get
![\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B72.0g%5Ctimes%202.09J%2Fg%5E0C%5Ctimes%20%280-%28-18%29%5E0C%5D%2B4.00mole%5Ctimes%206010J%2Fmole%2B%5B72.0g%5Ctimes%204.184J%2Fg%5E%29C%5Ctimes%20%2825-0%29%5E0C%5D)
(1 KJ = 1000 J)
Therefore, the enthalpy change is 34.3 kJ