Answer:
His measurements will be more precise if he uses a graduated cylinder that measures to the nearest tenth of a mL.
Explanation:
Answer:
B and D
Explanation:
it is B and D because in no gravity if you change the size of the ball it does not change anything. and if you swing fast it does not change anything but the speed. if you change the angle it is just the same modle but different angle
Explanation:
The chemical equation is as follows.

And, the given enthalpy is as follows.
;
= 102.5 kJ
Cl-Cl = 243 kJ/mol, O=O = 498 kJ/mol
Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.
102.5 = ![[(\frac{1}{2})x + 498] - [(2)(243)]](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29x%20%2B%20498%5D%20-%20%5B%282%29%28243%29%5D)
102.5 = 
102.5 - 12 = 
x = 181 kJ
Now, total bond enthalpy of per mole of ClO is calculated as follows.

x = ![[(\frac{1}{2})181 + (\frac{1}{2})498] - 243](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29181%20%2B%20%28%5Cfrac%7B1%7D%7B2%7D%29498%5D%20-%20243)
= 339.5 - 243
= 96.5 kJ
Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.
Answer:
1.2 × 10⁸ m
Explanation:
You want to convert metres to nanometres.
Recall that the multiplying prefix nano- means × 10⁻⁹, so
1 nm = 1 × 10⁻⁹ m
The conversion factor is either
or
.
You choose the one that has the desired units (nm) on top. Then

Answer:
16. Option c.
Explanation:
The reaction is:
C₈H₁₈ + O₂ → CO₂ + H₂O
Let's count, we have 8 C, 18 H and 2 O in reactant side
We have 1 C, 2 H and 3 O in product side
In order to balance the C, we add 8 to CO₂ and to balance the H, we add a 9 to H₂O
Now we have 8 C on both sides and 18 H On both sides. Finally we have 25 O on product side.
C₈H₁₈ + O₂ → 8CO₂ + 9H₂O
To balance the O in product side we must add 25/2, but as it is a rational number, we must multiply x2 to get an integer number (x2 in all the stoichiometry). The balanced reaction is :
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
with 16 C, 36 H and 50 O, on both sides