yes because its kind of like a seesaw if theres too much on one side its unbalance there has to be the same amount of atoms on each side
Answer: 2.17mol/L
Explanation:
C=n/V
n: 0.217mol
V: 100.0mL
1. Convert mL to L (100.0mL/1000 = 0.01L)
2. Put numbers into equation (C= 0.217/0.01L)
3. Molarity is 2.17mol/L
<u>Answer:</u> The
for the reaction is 51.8 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical equation for the reaction of carbon and water follows:

The intermediate balanced chemical reaction are:
(1)
( × 2)
(2)
( × 2)
(3)

The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B2%5Ctimes%20%5CDelta%20H_1%5D%2B%5B2%5Ctimes%20%5CDelta%20H_2%5D%2B%5B1%5Ctimes%20%28-%5CDelta%20H_3%29%5D)
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-393.7%29%29%2B%282%5Ctimes%20%28-285.9%29%29%2B%281%5Ctimes%20-%28-1411%29%29%5D%3D51.8kJ)
Hence, the
for the reaction is 51.8 kJ.
A possible cause of a large percentage of error in an
experiment where MgO is produced from the combustion of magnesium would be not all of the Mg has
completely reacted. <span>
I hope this helps and if you have any further questions, please don’t hesitate
to ask again. </span>
Radioactive decay => C = Co { e ^ (- kt) |
Data:
Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min
Time conversion: 4 hr 39 min = 4.65 hr
1) Replace the data in the equation to find k
C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}
=> k = ln {Co / C} / t = ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719
2) Use C / Co = 1/2 to find the hallf-life
C / Co = e ^ (-kt) => -kt = ln (C / Co)
=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k
t = ln(2) / 0.44719 = 1.55 hr.
Answer: 1.55 hr