Answer:
substitution is the best method or collecting like terms
Explanation:
Answer:
91383 J
Explanation:
The equation of the reaction can be represented as:
------>
Given that:
The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.
The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.
= 
where:
= enthalpy of reaction
= the difference in the heat capacities of the products and the reactants.
∴
= 
= ![1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'](https://tex.z-dn.net/?f=1%2891300%20J.mol%5E%7B-1%7D%20%29%20%2B%5Cint%5Climits%5E%7B435%7D_%7B298.15%7D%20%5B%7B%2829.86%29-%5Cfrac%7B1%7D%7B2%7D%2829.38%29-%5Cfrac%7B1%7D%7B2%7D29.13%7D%5DJ.K%5E%7B-1%7D.mol%5E%7B-1%7D%20%5C%2C%20dT%27)
= 91300 J + (0.605 J.K⁻¹)(435-298.15)K
= 91382.79 J
≅ 91383 J
40.0 g ( 1 mole ) --------------- 6.02x10²³ molecules
? ? --------------------------- 2.90x10²² molecules
mass = 2.90x10²² * 40.0 / 6.02x10²³
mass = 1.16x10²⁴ / 6.02x10²³
mass = 1.9269 g
hope this helps!
Answer:
C. The half-life of C-14 is about 40,000 years.
Explanation:
The only false statement from the options is that the half-life of C-14 is 40,000yrs.
The half-life of an isotope is the time it takes for half of a radioactive material to decay to half of its original amount. C-14 has an half-life of 5730yrs. This implies that during every 5730yrs, C-14 will reduce to half of its initial amount.
- All living organisms contain both stable C-12 and the unstable isotope of C-14
- The lower the C-14 compared to the C-12 ratio in an organism, the older it is.
