Answer:
Charle's Law
V1/T1=V2/T2
Explanation:
Know gas laws
- Hope this helped! Let me know if you need further explanation.
Answer:
a) 1,332 kJ are transferred
b) the temperature of the tree decreases 1.99 ºC
Explanation:
a) When a phase change occurs, the temperature remains constant for a pure substance, so, in this case, the heat transferred is:
Q = m*L, where m is the mass, and L is the latent heat, which for solidification of water is -333 kJ/kg:
Q = 4.00*(-333)
Q = -1332 kJ
So, 1,332 kJ are transferred.
b) If the amount of heat is transferred for the tree, and it has no phase change, it will receive a sensitive heat, which is:
Q = m*c*ΔT, where m is the mass, c is the specific heat, and ΔT the temperature change:
-1332 = 200*3.35*ΔT
670ΔT = -1332
ΔT = - 1.99 ºC
So the temperature of the tree decreases 1.99 ºC.
CaCO3 is the formula-- do you mean the name of it?
That would be calcium carbonate
If you mean the equation of formation, that would be:
(Ca^2+) + (CO3^2-) --> CaCO3
Aluminum hydroxide
can behave as a base and neutralize sulfuric acid
as in the following equation:
(Balanced)
(a)
. Thus the ratio between the number of moles of the two reactants available:
![n(\text{Al}(\text{OH})_3, \text{supplied}) / n(\text{H}_2\text{SO}_4, \text{supplied})\\= [m(\text{Al}(\text{OH})_3)/ M(\text{Al}(\text{OH})_3)] / [n(\text{H}_2\text{SO}_4) / M(\text{H}_2\text{SO}_4)]\\= [23.7 / (26.98 + 3 \times(16.00 + 1.008))]/[29.5 / (2 \times 1.008 + 32.07 + 4 \times 16.00)]\\\approx 1.01](https://tex.z-dn.net/?f=n%28%5Ctext%7BAl%7D%28%5Ctext%7BOH%7D%29_3%2C%20%5Ctext%7Bsupplied%7D%29%20%2F%20n%28%5Ctext%7BH%7D_2%5Ctext%7BSO%7D_4%2C%20%5Ctext%7Bsupplied%7D%29%5C%5C%3D%20%5Bm%28%5Ctext%7BAl%7D%28%5Ctext%7BOH%7D%29_3%29%2F%20M%28%5Ctext%7BAl%7D%28%5Ctext%7BOH%7D%29_3%29%5D%20%2F%20%5Bn%28%5Ctext%7BH%7D_2%5Ctext%7BSO%7D_4%29%20%2F%20M%28%5Ctext%7BH%7D_2%5Ctext%7BSO%7D_4%29%5D%5C%5C%3D%20%5B23.7%20%2F%20%2826.98%20%2B%203%20%5Ctimes%2816.00%20%2B%201.008%29%29%5D%2F%5B29.5%20%2F%20%282%20%5Ctimes%201.008%20%2B%2032.07%20%2B%204%20%5Ctimes%2016.00%29%5D%5C%5C%5Capprox%201.01)
The value of this ratio required to lead to a complete reaction is derived from coefficients found in the balanced equation:
![n(\text{Al}(\text{OH})_3, \text{theoretical}) / n(\text{H}_2\text{SO}_4, \text{theoretical}) = 2/3 \approx 0.667](https://tex.z-dn.net/?f=n%28%5Ctext%7BAl%7D%28%5Ctext%7BOH%7D%29_3%2C%20%5Ctext%7Btheoretical%7D%29%20%2F%20n%28%5Ctext%7BH%7D_2%5Ctext%7BSO%7D_4%2C%20%5Ctext%7Btheoretical%7D%29%20%3D%202%2F3%20%5Capprox%200.667)
The ratio for the complete reaction is smaller than that of the reactants available, indicating that the species represented on the numerator,
, is in excess while the one on the denominator,
, serves as the limiting reagent.
(b)
The quantity of water produced is dependent on the amount of limiting reactants available.
of sulfuric acid is supplied in this reaction as the limiting reagent.
moles of water molecules are produced for every
moles of sulfuric acid consumed. The reaction would thus give rise to
of water molecules, which have a mass of
.
(c)
![\text{Percentage Yield}\\= \text{Actual Yield} / \text{Theoretical Yield} \times 100 \; \%\\= 2.21 / 10.8 \times 100 \; \%\\= 20.4 \; \%](https://tex.z-dn.net/?f=%5Ctext%7BPercentage%20Yield%7D%5C%5C%3D%20%5Ctext%7BActual%20Yield%7D%20%2F%20%5Ctext%7BTheoretical%20Yield%7D%20%5Ctimes%20100%20%5C%3B%20%5C%25%5C%5C%3D%202.21%20%2F%2010.8%20%5Ctimes%20100%20%5C%3B%20%5C%25%5C%5C%3D%2020.4%20%5C%3B%20%5C%25)
(d)
The quantity of
, the reactant in excess, is dependent on the number of moles of this species consumed in the reaction and thus the quantity of the limiting reagent available. The consumption of every
moles of sulfuric acid, the limiting reagent, removes
moles of aluminum hydroxide
from the solution.
of sulfuric acid is initially available as previously stated such that
, or
, of
would be eventually consumed.
of
would thus be in excess by the end of the reaction process.