B, C and E. In radioactive decay unstable atoms stabilize by releasing energy.
Hi there! Let's solve this problem shall we!
⠀Volume = 10g
Mass = 2 mL
In this specific problem, they are asking us to find the <u><em>density </em></u>of the object. So,<u><em> using the information given to us</em></u> (volume and mass), let's solve the problem!
Now, if you remember, D = M ÷ V
So, let's fill in the blanks!
D = Our unknown value
M = 2mL
V = 10g
Here is the filled out formula:
D = M ÷ V
D = 2mL ÷ 10g
D = 5 g/mL
*Make sure you put the units for your final solution!*
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
The given reaction is:
C4H10 + O2 → CO2 + H2O
The above equation is not balanced due to the unequal distribution of atoms on either side of equation
# atoms Reactants # atoms products
C = 4 C = 1
H = 10 H = 2
O = 2 O = 3
In order to balance it, multiply C4H10 by 2, O2 by 13, CO2 by 8 and H2O by 10 to get:
2C4H10 + 13 O2 → 8CO2 + 10H2O
