For i: 33mL
For ii: 87-88mL
For iii:22.3mL
Answer:
because a fruit holds the seeds it needs to reproduce and continue living.
Answer : The correct option is, (C) 0.675 M
Explanation :
Using neutralization law,

where,
= concentration of
= 13.5 M
= concentration of diluted solution = ?
= volume of
= 25.0 ml = 0.0250 L
conversion used : (1 L = 1000 mL)
= volume of diluted solution = 0.500 L
Now put all the given values in the above law, we get the concentration of the diluted solution.


Therefore, the concentration of the diluted solution is 0.675 M
Here you are looking on the Free Body diagram of a net force of 0N in both the x and y-directions. the only ones that has that condition met is A and C.
<u>Answer:</u> The
for the reaction is -1052.8 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 given chemical reaction follows:

The intermediate balanced chemical reaction are:
(1)

(2)

The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B1%5Ctimes%20%5CDelta%20H_1%5D%2B%5B1%5Ctimes%20%28-%5CDelta%20H_2%29%5D)
Putting values in above equation, we get:

Hence, the
for the reaction is -1052.8 kJ.