Answer:
219.78 mL of the stock solution are needed
Explanation:
First, we take a look at the desired Al2(SO4)3 working solution. We are told that the we need 400 mL of an aqueous aluminum sulfate solution 1.0 M. Let's see how many moles of the compound we have in the desired volume:
1000 mL Al2(SO4)3 solution ----- 1 mole of Al2(SO4)3
400 mL Al2(SO4)3 solution ----- x = 0.4 moles of Al2(SO4)3
To reach the desired concentration in the working solution we need 0.4 moles of Al2(SO4)3 in 400 mL, so we calculate the volume of the stock solution needed to prepare the working solution:
1.82 moles Al2(SO4)3/L = 1.82 M → This is the molar concentration of the stock solution.
1.82 moles of Al2(SO4)3 ----- 1000 mL
0.4 moles of Al2(SO4)3 ----- x = 219.78 mL
So, if we take 219.78 mL of the 1.82 M stock solution, we put it in a graduated cylinder and we dilute it to 400 mL, we would obtain a 1.0 M Al2(SO4)3 solution.
Answer:
The correct answer will be "18.25 g".
Explanation:
The given values are:
Specific heat,
C = 0.45 J/g･°C
Heat involved,
q = 801 J
Temperature,
ΔT = 120.0°C-22.5°C
= 97.5°C
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
Answer:
CONVECTION-- Thermal energy is transferred from hot places to cold places by convection. Convection occurs when warmer areas of a liquid or gas rise to cooler areas in the liquid or gas. Cooler liquid or gas then takes the place of the warmer areas which have risen higher.
Explanation:
Answer:
2487.51.
Explanation:
As per Boyle's law temperature remaining constant the volume of an ideal gas is inversely proportional to its pressure.
pV= k
therefore, p1V1 = p2V2
here V1 = 25.3, p2 = 8.04mm Hg
pressure p1 = 790.5 mm Hg
this means that
25.3×790.5 = 8.04V2
⇒V2= 2487.51
Hence, the required volume is, 2487.51.
Answer:
Explanation:
There are three heat transfers involved.
heat from combustion of propane + heat gained by water + heat gained by calorimeter = 0
q₁ + q₂ + q₃ = 0
m₁ΔH + m₂C₂ΔT + C_calΔT = 0
Data:
m₁ = 2.1 g
m₂ = 280 g
Ti = 25.00 °C
T_f = 26.55 °C
Ccal = 92.3 J·°C⁻¹
Calculations:
Let's calculate the heats separately.
1. q₁
q₁ = 2.1 g × ΔH = 2.1ΔH g
2. q₂
ΔT = T_f - Ti = 26.55 °C - 25.00 °C = 1.55 °C
q₂ = 280 g × 4.184 J·°C⁻¹ × 1.55 °C = 1816 J
3. q₃
q₃ = 92.3 J·°C⁻¹ × 1.55 °C = 143.1 J
4. ΔH
