2Na(s) + 2H2O(l) = 2NaOH(aq) + H2(g)
Result the solution is a basic because .. in there added a strong base.. naoh
Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=![2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}](https://tex.z-dn.net/?f=2%2AH_%7BNO%7D%20%2BH_%7BCl_%7B2%7D%20%7D-2%2AH_%7BNOCl%7D)
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 - ![H_{NOCl}](https://tex.z-dn.net/?f=H_%7BNOCl%7D)
Solving
-
=75.5 kJ/mol - 2*90.25 kJ/mol
-
=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>
Bone age : 22,920 years
<h3>Further explanation</h3>
Given
Nt = 2.5 g C-14
No = 40 g
half-life = 5730 years
Required
time of decay
Solution
General formulas used in decay:
![\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}](https://tex.z-dn.net/?f=%5Clarge%7B%5Cboxed%7B%5Cbold%7BN_t%3DN_0%28%5Cdfrac%7B1%7D%7B2%7D%29%5E%7Bt%2Ft%5Cfrac%7B1%7D%7B2%7D%20%7D%7D%7D)
t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Input the value :
![\tt 2.5=40.\dfrac{1}{2}^{t/5730}\\\\\dfrac{2.5}{40}=\dfrac{1}{2}^{t/5730}\\\\(\dfrac{1}{2})^4=\dfrac{1}{2}^{t/5730}\\\\4=t/5730\rightarrow t=22920~years](https://tex.z-dn.net/?f=%5Ctt%202.5%3D40.%5Cdfrac%7B1%7D%7B2%7D%5E%7Bt%2F5730%7D%5C%5C%5C%5C%5Cdfrac%7B2.5%7D%7B40%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5E%7Bt%2F5730%7D%5C%5C%5C%5C%28%5Cdfrac%7B1%7D%7B2%7D%29%5E4%3D%5Cdfrac%7B1%7D%7B2%7D%5E%7Bt%2F5730%7D%5C%5C%5C%5C4%3Dt%2F5730%5Crightarrow%20t%3D22920~years)