02(g) = 0 kj/mol
<span>CO2 (g) = -393.5 kj/mol </span>
<span>H20(g) = -241.8 kj/mol </span>
<span>H total = -5094 kJ
</span>5094kJ = [8(-393.5) + 9(-241.8)] - [X + 12.5(0)]
<span>-5094 kJ = [-3148 + (-2176.2)] - [x + 0] </span>
<span>-5094 kJ = -5324.2 - x </span>
<span>add -5324.2 to -5094 </span>
<span>to get +230.2 = -x </span>
<span>move the negative to the other side </span>
<span>and you get -230 kj/mol</span>
Answer:
D.
Explanation:
-log(1.0x10^-5) = pH
pH + pOH = 14 (rearrange it)
OH- = 10^-pOH = 1.0 x 10^-9
- Hope that helped! Let me know if you need further explantion.
Hey there!
C₉H₂O + O₂ → CO₂ + H₂O
First let's balance the C.
There's 9 on the left and 1 on the right. So, let's add a coefficient of 9 in front of CO₂.
C₉H₂O + O₂ → 9CO₂ + H₂O
Next let's balance the H.
There's 2 on the left and 2 on the right. This means it's already balanced.
C₉H₂O + O₂ → 9CO₂ + H₂O
Lastly, let's balance the O.
There's 3 on the left and 19 on the right. So, let's add a coefficient of 9 in front of O₂.
C₉H₂O + 9O₂ → 9CO₂ + H₂O
This is our final balanced equation.
Hope this helps!
Alkaline Earth Metals are the elements located in the second period from the left of the periodic table. These elements lose two electrons to form the stable octet when forming an ionic bond, resulting in a net charge of +2. Because they’re trying to get rid of those electrons to get to the stable octet, it’s easy to remove them - this means that the ionization energy of these elements is relatively low. Finally, since they’re looking to get rid of electrons, they certainly aren’t trying to gain any, meaning that their electronegativity is relatively low.
The correct answers are A and D.
