Answer:
8.9 mg/l
Explanation: Temp doesnt matter so throw that out automatically then your equation is;
S1/P1=S2/P2
We are looking for S2 and that equation is;
S2=S1*P2/P1 and that is S2=22.25*1/2.5
A little bit of simple math and you get your answer: 8.9 mg/l
The amount in grams of Al₂O₃ produced is approximately 6.80 g.
Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:
AL + O₂ → Al₂O₃
Let's balance it
4AL + 3O₂ → 2Al₂O₃
4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,
Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,
Atomic mass of AL = 27 g
Molar mass of Al₂O₃ = 101.96 g/mol
4(27 g) of AL gives 2(101.96 g) of Al₂O₃
3.6 g of AL will give ?
cross multiply
mass of Al₂O₃ produced = 3.6 × 203.92 / 108 = 734.112 / 108 = 6.797
mass of Al₂O₃ produced = 6.80 g.
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B. It is the heat required to change a gram of substance from solid to liquid
Answer and Explanation:
Calorie is the unit of heat energy . There are 2 units with the same name 'calorie' which is widely used.
'The amount of heat energy required to increase the temperature of 1 gram of water by mass by 
or 1 K is known as small calorie or gram calorie'.
Another one is large calorie which can be defined as :
'The amount of heat energy required to make arise in temperature of water 1 kg by mass by or 1 K is known as large calorie or kilcalorie and is represented as Cal or kcal'.
After the adoption of SI system, thee units of the metric system cal, C or kilocal are considered deprecated or obsolete with the SI unit for heat energy as 'joule or J'
1 cal = 4.184 J
1C or 1 kilocal = 4184 J
Calorimeter constant:
Calorimeter constant, represented as '
' is used to quantify the heat capacity or the amount of heat of a calorimeter.
It can be calculated by ther given formula:
where,
= corresponding temperature change
= enthalpy change
Its unit is J/K or J/1^{\circ}C[/tex] which can be convertyed to cal/1^{\circ}C[/tex] by dividing the calorimeter constant by 4.184 or 4184 accordingly.
