Answer:
3m/s²
Explanation:
Given parameters:
Mass of object = 3.2kg
Force to the right = 16.3N
Force to the left = 6.7N
Unknown:
Acceleration of the object = ?
Solution:
To solve this problem, we use newtons second law of motion;
Net force = mass x acceleration
Net force on object = Force to the right - Force to the left
Net force = 16.3N - 6.7N = 9.6N
So;
9.6 = 3.2 x a
a =
= 3m/s²
<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.
7 would be correct because I divide it
Answer:
sorry i'm not a scientist but i think it's C
Explanation:
sorry again if i get it wrong for u :(
Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2