Answer:
Static electricity is a type of energy that is produced by friction.
Explanation:
Static electricity may be the result when there is an imbalance between the negative charges and the positive charges in the surroundings. They are non contact forces. They pull of push without actually touching the body. The charged particles interact by pulling or pushing the uncharged particles.
These charges are released when they come closer to some uncharged or other charged particles like the electrostatic charges. It is the charge build up in an area.
Answer:
Volume of sample after droping into the ocean=0.0234L
Explanation:
As given in the question that gas is idealso we can use ideal gas equation to solve this;
Assuming that temperature is constant;
Lets
and
are the initial gas parameter before dropping into the ocean
and
and
are the final gas parameter after dropping into the ocean
according to boyle 's law pressure is inversly proportional to the volume at constant temperature.
hence,

P1=1 atm
V1=1.87L
P2=80atm
V2=?
After putting all values we get;
V2=0.0234L
Volume of sample after droping into the ocean=0.0234L
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
The heat lost by the metal should be equal to the heat
gained by the water. We know that the heat capacity of water is simply 4.186 J
/ g °C. Therefore:
100 g * 4.186 J / g °C * (31°C – 25.1°C) = 28.2 g * Cp *
(95.2°C - 31°C)
<span>Cp = 1.36 J / g °C</span>
Answer:
is the formula for the limiting reagent.
Mass of silver chloride produced is 71.8 g.
Explanation:

Moles of silver nitrate = 0.500 mol
Moles of copper(II) chloride = 0.285 mol
According to reaction, 2 moles of silver nitrate reacts with 1 mole of copper chloride , then 0.500 mole of silver nitrate will react with :
of copper(II) chloride
As we can see that moles of copper(II) chloride will be reacting is 0.250 mol less than present moles of copper (II) chloride ,so this means that silver nitrate is limiting reagent.
And moles of silver chloride to be formed will depend upon silver nitrate.
According to reaction, 2 moles of silver nitrate gives 2 moles of silver chloride , then 0.500 mole of silver nitrate will give :
of silver chloride
Mass of silver chloride produced:
0.500 mol × 143.5 g/mol = 71.8 g