Answer:
0.11mole
Explanation:
Let us assume that the condition is at standard temperature and pressure(STP);
Given parameters:
Volume of water = 2.45L
Unknown:
Number of moles found in this volume of water = ?
Solution;
At STP;
Number of moles =
Input the parameters and solve;
Number of moles of water = = 0.11mole
The number of moles of water found is 0.11mole
The answer would be B because you can go and find evidence and facts to conduct an investigation on that.
Answer:
The correct answer is -1085 KJ/mol
Explanation:
To calculate the formation enthalphy of a compound by knowing its lattice energy, you have to draw the Born-Haber cycle step by step until you obtain each element in its gaseous ions. Find attached the correspondent Born-Haber cycle.
In the cycle, Mg(s) is sublimated (ΔHsub= 150 KJ/mol) to Mg(g) and then atoms are ionizated twice (first ionization: ΔH1PI= 735 KJ/mol, second ionization= 1445 KJ/mol) to give the magnesium ions in gaseous state.
By other hand, the covalent bonds in F₂(g) are broken into 2 F(g) (Edis= 154 KJ/mol) and then they are ionizated to give the fluor ions in gaseous state 2 F⁻(g) (2 x ΔHafinity=-328 KJ/mol). The ions together form the solid by lattice energy (ΔElat=-2913 KJ/mol).
The formation enthalphy of MgF₂ is:
ΔHºf= ΔHsub + Edis + ΔH1PI + ΔH2PI + (2 x ΔHaffinity) + ΔElat
ΔHºf= 150 KJ/mol + 154 KJ/mol + 735 KJ/mol + 1445 KJ/mol + (2 x (-328 KJ/mol) + (-2913 KJ/mol).
ΔHºf= -1085 KJ/mol
Answer:
Explanation:
The changes in properties from metals to non-metals on a periodic table can be measured and determined by the metallicity or electropositivity of elements.
Metallicity is a measure of the tendency of atoms of an element to lose electrons.
a.
Down a periodic group, metallicity increases.
b.
Across a period from left to right electropositivity or metallicity decreases.
Metals are found in the left part of periodic table and the most reactive metal sits in the lower left corner. Non-metals are towards the right side of the table.
Answer:
4.5moles
Explanation:
First, let us balance the equation given from the question. This is illustrated below:
KClO3 —> KCl + O2
There are 2 atoms of O on the right side and 3 atoms on the left. It can be balance by putting 2 in front of KClO3 and 3 in of O2 as shown below
2KClO3 —> KCl + 3O2
Now, we have 2 atoms each of K and Cl on the left side and 1atom each of K and Cl on the right. It can be balance by putting 2 in front of KCl as shown below:
2KClO3 —> 2KCl + 3O2
Now the equation is balanced.
From the balanced equation,
2 moles of KClO3 produced 3 moles of O2.
Therefore, 3 moles of KClO3 will produce = (3 x 3) /2 = 4.5moles of O2.
Therefore 3 moles of KClO3 will produce 4.5 moles of O2