Answer: 0.9375 g
Explanation:
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of 
solution = 0.75 M
Volume of solution = 25.0 mL = 0.025 L
Putting values in equation 1, we get:
According to stoichiometry :
2 moles of require = 1 mole of 
Thus 0.01875 moles of will require=
of
Mass of 
Thus 0.9375 g of is required to react with 25.0 ml of 0.75 M HCl
Answer:
it was Millikan. He conducted the oil drop experiment. Thomson determined the electon charge not the quantity. Rutherford used the gold foil experiment to find positive charge and that most of the atom is empty space. Dalton proposed that matter was made of small particles called atoms but that was a concept already proposed by ancient greeks. Dalton also proposed the atomic theory.
Enthalpy is a thermodynamic quantity that describes the heat content of a system, that can not be measured directly. That's why we measure change in enthaply, measured in the units joules. The statement that e<span>nthalpy change depends on the rate at which a substance is heated or cooled is false. Enthalpy change depends only on the following factors:
-</span><span>physical state of reactants and products
- quantity of reactants</span><span>
- allotropic modifications
- temperature and pressure</span><span>
</span>
Answer: The reaction is exothermic reaction as the energy of products is less than the energy of reactants.
Explanation: Exothermic reactions are defined as the reactions in which energy of the product is less than the energy of the reactants. The total energy is released in the form of heat and 
for the reaction comes out to be negative.
Labeling of the parts in the diagram:
A represents the activation energy which is the energy required by reactants to cross the energy barrier to get converted to products.
E represents the potential energy of the reactants.
B represents the activated complex.
D represents the potential energy of the products.
C represents the total enthalpy change of the reaction, which comes out to be negative for exothermic reactions.
Answer:
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