This question can be simply solved by using heat formula,
Q = mCΔT
Q = heat energy (J)
m = Mass (kg)
C = Specific heat capacity (J / kg K)
ΔT = Temperature change (K)
when water freezes, it produces ice at 0°C (273 K)
hence the temperature change is 25 K (298 K - 273 K)
C for water is 4186 J / kg K or 4.186 J / g K
By applying the equation,
Q = 456 g x 4.186 J / g K x 25 K
= 47720.4 J
= 47.72 kJ
hence 47.72 kJ of heat energy should be removed.
Answer:
<u>Kinetic particle theory</u>
Arrangement and motion of solid particles
-> Solid particles are packed closely with each other in an orderly manner. They vibrate vigorously in their fixed positions.
Arrangement and motion of liquid particles
-> Liquid particles are packed less closely with each other as compared to solid particles in a disorderly manner. They move around in a random motion; sliding past each other.
Answer:
We have to add 9.82 grams of calcium acetate
Explanation:
Step 1: Data given
Molarity of the calcium acetate solution = 0.207 M
Volume = 300 mL = 0.300 L
Molar mass calcium acetate = 158.17 g/mol
Step 2: Calculate moles calcium acetate
Moles calcium acetate = molarity * volume
Moles calcium acetate = 0.207 M * 0.300 L
Moles calcium acetate = 0.0621 moles
Step 3: Calculate mass calcium acetate
Mass calcium acetate = moles * molar mass
Mass calcium acetate = 0.0621 moles * 158.17 g/mol
Mass calcium acetate = 9.82 grams
We have to add 9.82 grams of calcium acetate
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
The charge of a Rb ion would be +1
