Answer:
Hope this helps
Explanation:
https://seagrant.whoi.edu/wp-content/uploads/2018/05/ESTIMATING-POPULATION-SIZE-1.pdf
Answer:
The density of copper is 0.5 g/mL
Explanation:
Given data:
Mass of copper = 6 g
Volume of copper = 12 mL
Density of copper = ?
Solution:
Formula:
d = m/v
d = density
m = mass
v = volume
d = 6 g/ 12 mL
d = 0.5 g/mL
Thus, the density of copper is 0.5 g/mL
Rutherford theorized that atoms have their charge concentrated in a very small nucleus.
This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).
Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.
That is different from Plum Pudding model, because it shows that most of the atom is empty space.
According to Rutherford model of the atom:
1) Atoms have their charge concentrated in a very small nucleus.
2) Major space in an atom is empty.
3) Atoms nucleus is surrounded by negatively charged particles called electrons.
4) An atom is electrically neutral.
Answer:
A cation is formed when a metal ion loses a valence electron while an anion is formed when a non-metal gains a valence electron. They both achieve a more stable electronic configuration through this exchange.
Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
