Answer:
1) 0.0625 g.
2) 0.0125 g.
Explanation:
<em>1) A solution of NaOH has a concentration of 25.00% by mass. What mass of NaOH is present in 0.250 g of this solution?</em>
mass% of NaOH = [(mass of NaOH)/(mass of solution)] x 100.
mass% of NaOH = 25.0%, mass of NaOH = ??? g, mass of solution = 0.250 g.
∴ mass of NaOH = (mass% of NaOH)(mass of solution)/100 = (25.0%)(0.250 g)/100 = 0.0625 g.
<em>2) What mass of NaOH must be added to the solution to increase the concentration to 30.00% by mass?</em>
We can use the relation:
mass% of NaOH = [(mass of NaOH)/(mass of solution)] x 100.
mass% of NaOH = 30.0%, mass of NaOH = ??? g, mass of solution = 0.250 g.
∴ mass of NaOH = (mass% of NaOH)(mass of solution)/100 = (30.0%)(0.250 g)/100 = 0.075 g.
∴ The mass of NaOH should be added = 0.075 - 0.0625 = 0.0125 g.
Answer:
The valence shell has higher energy than other occupied shells
Explanation:
According to Bohr's model of the atom, he suggested that the extranuclear part consists of electrons in specific spherical orbits around the nucleus.
His model suggests that the electron can move round the nucleus in certain permissible orbits or energy levels. The ground state is the lowest energy state available to the electron. The excited state is any level higher than the ground state.
The valence electrons are in the outermost shell of an atom. These electrons are of the highest energy levels in the atom
THANKS
Explanation:
Below is an attachment containing the solution.
Answer:
True
Explanation:
The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).
For the reaction;
N2(g) + O2(g) = 2NO(g)
Kp = pNO²/ pN₂pO₂; (No units)
where;
pNO is the partial pressure of NO;
pN₂ is the partial pressure of nitrogen
pO₂ is the partial pressure of Oxygen
The equilibrium constant Kp is deduced from the balanced chemical equation for a reversible reaction, NOT experimental data as is the case for rate expressions in kinetics.
Kp changes with temperature considerably changing the position of an equilibrium, and, at a constant temperature, and therefore constant K, the position of an equilibrium can change significantly depending on relative concentrations/pressures of 'reactants' and 'products'.
