Answer:
If two tectonic plates collide, they form a convergent plate boundary. Usually, one of the converging plates will move beneath the other, a process known as subduction. ... As the sinking plate moves deeper into the mantle, fluids are released from the rock causing the overlying mantle to partially melt.
This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.
Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:

Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:

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Nun nun hmm I’m ummm I don’t wanna see how much
Answer:
Explanation:
<u>1. Chemical quation</u>
The reaction of aluminium, sodium hydroxide and water is represented by the balanced chemical equation:
- 2Al(s) + 2NaOH(s) + 6H₂O(l) → 2Na[Al(OH)₄] (aq) + 3H₂(g) ↑
The coefficients of each reactant and product give the theoretical mole ratios.
To find the limiting reactant you compare the theoretical ratios with the ratio of the available substaces.
<u>2. Theoretical mole ratio:</u>
- 2 mol Al : 2 mol NaOH : 6 mol H₂O
Equivalent to
- 1 mol Al : 1 mol NaOH : 3 mol H₂O
<u>3. Actual ratio</u>
a) Convert each mass to number of moles
Formula:
- number of moles = mass in grams / molar mass
Al:
- molar mass = atomic mass = 26.982g/mol
- number of moles = 51.0g / 26.982g/mol = 1.89 mol
NaOH:
- number of moles = 84.1g / 39.997g/mol = 2.10 mol
H₂O:
- number of moles = 25.0g / 18.015g/mol = 1.39 mol
Divide all the mole amounts by the least number:
- Al: 1.89/1.39 = 1.36
- NaOH: 2.10 = 1.52
- H₂O: 1.39 = 1.00
- 1.36 mol Al : 1.52 mol NaOH : 1.00 mol H₂O
<u>4. Comparison</u>
<u />
Theoretical ratio:
- 1 mol Al : 1 mol NaOH : 3 mol H₂O
Actual ratio:
- 1.36 mol Al : 1.52 mol NaOH : 1.00 mol H₂O
Multiply by 3:
- 4.08 mol Al : 4.56 mol NaOH : 3.00 mol H₂O
Now, yo can see that the first two are in excess with respect the third one, making that the water consumes first, before any of the other two consumes. Therefore, the limiting reactant is the water.
Answer:- 
Solution:- First of all we calculate the heat absorbed or released when the solute is added to the solvent. Here the solute is LiCl and the solvent is water.
To calculate the heat absorbed or released we use the formula:

q = heat absorbed or released
m = mass of solution
s = specific heat capacity
and
= change in temperature
mass of solution = mass of solute + mass of solvent
mass of solution = 5.00 g + 100.0 g = 105.0 g
(note:- density of pure water is 1 g per mL so the mass is same as its volume)
= 33.0 - 23.0 = 10.0 degree C
s = 
Let's plug in the values in the formula and calculate q.
q = 
q = 4389 J
To calculate the enthalpy of solution that is
we convert q to kJ and divide by the moles of solute.
moles of solute = 
= 0.118 moles
q =
= 4.389 kJ

= 
Since the heat is released which is also clear from the rise in temperature of the solution, the sign of enthapy of solution will be negative.
So, 