Answer:
0.020 moles of
can be formed
Explanation:
1. First determine the number of moles of LiOH.
Molarity is given by the following expression:

Solving for moles of solute:
moles of solute = M * Liters of solution
Converting 175.0mL to L:

Replacing values:
moles of solute = 0.227M*0.175L
moles of solute = 0.040
Therefore there are 0.040 moles of LiOH
2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of
produced:

As the problem says that there are excess of
, the limiting reagent is the LiOH.
can be formed
The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ <span>joule = 1 eV
Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules
From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s
Answer:

Explanation:
Hello there!
In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

Thus, we solve for the molarity of the acid to obtain:

Regards!
0.17 M is the is the molal concentration of this solution
Explanation:
Data given:
freezing point of glucose solution = -0.325 degree celsius
molal concentration of the solution =?
solution is of glucose=?
atomic mass of glucose = 180.01 grams/mole
freezing point of glucose = 146 degrees
freezing point of water = 0 degrees
Kf of glucose = 1.86 °C
ΔT = (freezing point of solvent) - (freezing point of solution)
ΔT = 0.325 degree celsius
molality =?
ΔT = Kfm
rearranging the equation:
m = 
m= 0.17 M
molal concentration of the glucose solution is 0.17 M