Supercooling<span>, a state where liquids do not solidify even below their normal freezing point. Means sometimes we have liquid water below 0 degree C.</span>
Answer:
0.48 V
Explanation:
Zn(s) ------------> Zn^2+(aq) + 2e. Oxidation half equation (-0.76V)
Co^2+(aq) + 2e-----------> Co(s). Reduction half equation (-0.28)
Zn(s) + Co^2+(aq) -------------> Zn^2+(aq) + Co(s) overall redox equation
Zinc is the anode while cobalt is the cathode.
E°cell= E°cathode - E°anode
E°cell= -0.28-(-0.76)= 0.48 V
For this question, assume that you have 1 compound. This compound is divided in half once, so you are left with 0.5. That 0.5 that remains is divided in half again, this is the second half-life, and you are left with 0.25. The final half life involves dividing 0.25 in half, which means you are left with 0.125. For the answer to make sense, you need to know your conversions between decimals and fractions. To make it simple, if you have 0.125 and you times it by 8, you are left with your initial value of 1. Therefore, after three half-lives, you are left with 1/8th of the compound.
Answer:
kJ/mol
Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.
The standard enthalpy of formation of HCl is calculated as:

→ 
Standard Enthalpy of formation for the other compounds are:
Calcium Hydroxide:
-1002.82 kJ/mol
Calcium chloride:
-795.8 kJ/mol
Water:
-285.83 kJ/mol
Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.
Calculating:
![-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]](https://tex.z-dn.net/?f=-17.2%3D%5B-795.8%2B2%28285.85%29%5D-%5B-1002.82%2B2%5CDelta%20H%5D)



So, the standard enthalpy of formation of HCl is -173.72 kJ/mol
Answer:
0.098 moles
Explanation:
Let y represent the number of moles present
1 mole of Ba(OH)₂ contains 2 moles of OH- ions.
Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.
To get the y moles, we then do cross multiplication
1 mole * y mole = 2 moles * 0.049 mole
y mole = 2 * 0.049 / 1
y mole = 0.098 moles of OH- ions.
1 mole of OH- can neutralize 1 mole of H+
Therefore, 0.098 moles of HNO₃ are present.