<span>NaCl
First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound
Sodium = 22.989769
Chlorine = 35.453
Silver = 107.8682
Nitrogen = 14.0067
Oxygen = 15.999
Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results
For NaCl
22.989769 + 35.453 = 58.44277
For AgNO3
107.8682 + 14.0067 + 3 * 15.999 = 169.8719
Now calculate how many moles of each substance by dividing the total mass by the molar mass
For NaCl
4.00 g / 58.44277 g/mol = 0.068443 mol
For AgNO3
10.00 g / 169.8719 g/mol = 0.058868
Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>
A reaction in which Oxygen (O₂) is produced from Mercury Oxide (HgO) would be a decomposition reaction.
2HgO → 2Hg + O₂
If 250g of O₂ is needed to be produced,
then the moles of oxygen needed to be produced = 250g ÷ 32 g/mol
= 7.8125 mol
Now, the mole ratio of Oxygen to Mercury Oxide is 1 : 2
∴ if the moles of oxygen = 7.8125 mol
then the moles of mercury oxide = 7.8125 mol × 2
= 15.625 mol
Thus the number moles of HgO needed to produce 250.0 g of O₂ is 15.625 mol
Puberty is what is correct I believe
For this problem, the solution is exhibiting some colligative properties since the solute in the solution interferes with some of the properties of the solvent. We use equation for the boiling point elevation for this problem. We do as follows:
<span>
ΔT(boiling point) = (Kb)mi
</span>ΔT(boiling point) = (0.512)(1.3/2.0)(2)
ΔT(boiling point) = 0.67 degrees Celsius
<span>
T(boiling point) = 100 + 0.67 = 100.67 degrees Celsius</span>
Answer:
The empty space between the atomic cloud of an atom and its nucleus is just that: empty space, or vacuum. ... Electrons are thus 'spread out' quite a bit in their orbits about the nucleus. In fact, the wave-functions for electrons in s-orbitals about a nucleus actually extend all the way down into the nucleus itself.
