Start by adding the numbers then divide
The final volume V₂=4.962 L
<h3>Further explanation</h3>
Given
T₁=20 + 273 = 293 K
P₁= 1 atm
V₁ = 4 L
T₂=100+273 = 373 K
P₂=780 torr=1,02632 atm
Required
The final volume
Solution
Combined gas law :
P₁V₁/T₁=P₂V₂/T₂
Input the value :
V₂=(P₁V₁T₂)/(P₂T₁)
V₂=(1 x 4 x 373)/(1.02632 x 293)
V₂=4.962 L
<span>1 ml of water weighs 1 gram so 1 liter (1000 ml) weighs 1000 grams. A 3% solution (3% = 0.03) of hydrogen peroxide (w/v) would contain 1000 grams x 0.03 or 30 grams. The chemical formula of hydrogen peroxide is H2O2 and a mole weighs 34.0147 grams/mole. So 30 grams of H2O2 divided by 34.0147 grams/mole equals 0.88 moles of H2O2. The concentration of a 3% (w/v) hydrogen peroxide solution therefore contains 30 grams of H202 (or 0.88 moles of H202) per in a liter of water (or 1000 grams H20) would thus be 0.88 moles H2O2 per liter (0.88 moles H2O2/l) .</span>
This separation technique is a 4-step procedure. First, add H₂SO₄ to the solution. Because of common ion effect, BaSO₄ will not react, only Mg(OH)₂.
Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2 H₂O
The aqueous solution will now contain MgSO₄ and BaSO₄. Unlike BaSO₄, MgSO₄ is soluble in water. So, you filter out the solution. You can set aside the BaSO₄ on the filter paper. To retrieve Mg(OH)₂, add NaOH.
MgSO₄ + 2 NaOH = Mg(OH)₂ + Na₂SO₄
Na₂SO₄ is soluble in water, while Mg(OH)₂ is not. Filter this solution again. The Mg(OH)₂ is retrieved in solid form on the filter paper.
Answer: 90.04°C
Explanation: <u>Calorimeter</u> is a device measures the amount of heat of a chemical or physical process. An ideal calorimeter is one that is well-insulated, i.e., prevent the transfer of heat between the calorimeter and its surroundings. So, the net heat change inside the calorimeter is zero:
Rearraging, it can be written as
showing that the heat gained by Substance 1 is equal to the energy lost by Substance 2.
In our case, water is gaining heat, because its temperature has risen and so, brass is losing energy:
Calculating:
![m_{w}.c_{w}.\Delta T=-[m_{b}.c_{b}.\Delta T]](https://tex.z-dn.net/?f=m_%7Bw%7D.c_%7Bw%7D.%5CDelta%20T%3D-%5Bm_%7Bb%7D.c_%7Bb%7D.%5CDelta%20T%5D)
![100.4.18.(18.4-15)=-[52.9.0.375.(18.4-T)]](https://tex.z-dn.net/?f=100.4.18.%2818.4-15%29%3D-%5B52.9.0.375.%2818.4-T%29%5D)
Note: final temperature is the same as the substances are in thermal equilibrium.
Solving:
418(3.4)= - 365.01 + 19.8375T
19.8375T = 1786.21
T = 90.04
The initial temperature for the sample of brass was 90.04°.
