Answer:
146 kJ
Explanation:
There are two heat flows in this question.
Heat lost on cooling + heat lost on solidifying = 0
q₁ + q₂ = 0
mCΔT + nΔHsol = 0
Data:
m = 575 g
C = 0.449 J·K⁻¹g⁻¹
T_i = 1825 K
T_f = 1811 K
ΔHsol = -13.8 kJ·mol⁻¹
Calculations:
(a) Heat lost on cooling
ΔT = T_f - T_i = 1811 K - 1825 K = -14 K
q₁ = mCΔT = 575 g × 0.449 J·K⁻¹g⁻¹ × (-14 K) = -361 J = -3.61 kJ
(b) Heat lost on solidifying
(c) Total heat lost
q = q₁ + q₂ = -3.61 kJ - 142.1 kJ = -146 kJ
The heat lost was 146 kJ.
Answer:

Explanation:
2Al₂O₃ ⟶ 4Al +3O₂
n/mol: 1.26
The molar ratio is 4 mol Al:2 mol Al₂O₃.

Answer:
3Ba(OH)₂ + 2H₃PO₄ → Ba₃(PO₄)₂ + 6H₂O
Explanation:
Neutralization reaction:
When an acid and base react they form water and salt.The reaction is called neutralization reaction.
General equation:
HX + BOH → BX + H₂O
From given reaction equations only option C is correct because only this reaction gives salt and water.
Chemical equation:
3Ba(OH)₂ + 2H₃PO₄ → Ba₃(PO₄)₂ + 6H₂O
Barium hydroxide is acting as a base. H₃PO₄ is an acid. Both these reactants react to form salt which is barium phosphate and water.
Other options are incorrect because non of these reaction produced salt and water. So option A is correct.
Hi!
The reason for reviewing and replicating work of other scientists is influenced by two factors listed in the answer:
1. An experiment may have had errors that the scientist did not recognize. <u>(unintentional) </u>
2. The results of individual scientists may be influenced by bias. <u>(intentional)</u>
<em>Personal motivations may often influence a scientist to manipulate the results to best suit his/her hypothesis, consequently leading to scientific fraud. </em>
While other options listed are elements of the scientific method, they are not the reasons behind the review and replication of a scientist's work by others.
Charles law states that for a fixed mass of gas at constant pressure the volume of the gas is directly proportional to the temperature in Kelvin scale
v1/T1 = v2/T2
Where v1 is volume and T1 is temperature at the first instance
v2 is volume and T2 is temperature at the second instance
Substituting the values in the equation
0.5 L / 203 K = v2 / 273 K
v2 = 0.672 L
The volume of Air after heating is 0.672 L