<u>Answer:</u> The 
for the reaction is -1835 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) 
( × 4)
(2) 
The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B4%5Ctimes%20%28-%5CDelta%20H_1%29%5D%2B%5B1%5Ctimes%20%5CDelta%20H_2%5D)
Putting values in above equation, we get:
Hence, the for the reaction is -1835 kJ.
Hello!
Yes, it needs oxygen to metabolize sugars that, in the end, produce carbon dioxide (CO2).
Answer:
This is a pretty straightforward example of how an ideal gas law problem looks like.
Your strategy here will be to use the ideal gas law to find the pressure of the gas, but not before making sure that the units given to you match those used by the universal gas constant.
So, the ideal gas law equation looks like this
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
P
V
=
n
R
T
a
a
∣
∣
−−−−−−−−−−−−−−−
Here you have
P
- the pressure of the gas
V
- the volume it occupies
n
- the number of moles of gas
R
- the universal gas constant, usually given as
0.0821
atm
⋅
L
mol
⋅
K
T
- the absolute temperature of the gas
Take a look at the units given to you for the volume and temperature of the gas and compare them with the ones used in the expression of
R
.
a
a
a
a
a
a
a
a
a
a
a
Need
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
Have
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
Liters, L
a
a
a
a
a
a
a
a
a
a
a
a
a
Liters, L
a
a
a
a
a
a
a
a
a
a
a
√
a
a
a
a
a
a
a
Kelvin, K
a
a
a
a
a
a
a
a
a
a
a
a
Celsius,
∘
C
a
a
a
a
a
a
a
a
a
×
Notice that the temperature of the gas must be expressed in Kelvin in order to work, so make sure that you convert it before plugging it into the ideal gas law equation
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
T
[
K
]
=
t
[
∘
C
]
+
273.15
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−
Rearrange the ideal gas law equation to solve for
P
P
V
=
n
R
T
⇒
P
=
n
R
T
V
Plug in your values to find
P
=
0.325
moles
⋅
0.0821
atm
⋅
L
mol
⋅
K
⋅
(
35
+
273.15
)
K
4.08
L
P
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
2.0 atm
a
a
∣
∣
−−−−−−−−−−−
The answer is rounded to two sig figs, the number of sig figs you have for the temperature of the gas.
Answer:
group # = # valence electrons
group 2= 2 valence electrons
Explanation:
Mg is in group #2 and subsequently has 2 valence electrons (electrons on the outermost shell)
OH - REFERS TO THAT OT IS ION OF BASE IF THE SOLUTION OF BASE HAS PH 6.10 SO IT IS ACIDIC SO IT IS NOT POSSIBLE THAT SOLUTION OF OH- IS PH 6.10
