Answer:
82500000000000000000000000
Explanation:
This is the only answer I can come up with.
Molar solubility is number of moles of the solute that can be dissolved per liter of solution before the solution becomes saturated.
The molar solubility of lead(ii) chloride with ksp value of 2.4 × 10e4 can be solve as:
Ksp = s2 = 2.4 × 10e4
s2 = 2.4 × 10e4
s = √(2.4 × 10e4)
s = 154.9 mol/L
Answer:
The answer to the question above is
The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C = 29.36 kJ
Explanation:
The given variables are
ΔHfus = 7.27 kJ/mol
Cliq = 2.16 J/g°C
Cgas = 1.29 J/g°C
Csol = 1.65 J/g°C
Tmelting = -95.0°C.
Initial temperature = -154.0°C
Final temperature = -42.0°C?
Mass of acetone = 87.1 g
Molar mass of acetone = 58.08 g/mol
Solution
Heat required to raise the temperature of solid acetone from -154 °C to -95 °C or 59 °C is given by
H = mCsolT = 87.1 g* 1.65 J/g°C* 59 °C = 8479.185 J
Heat required to melt the acetone at -95 °C = ΔHfus*number of moles =
But number of moles = mass÷(molar mass) = 87.1÷58.08 = 1.5
Heat required to melt the acetone at -95 °C =1.5 moles*7.27 kJ/mol = 10.905 kJ
The heat required to raise the temperature to -42 degrees is
H = m*Cliq*T = 87.1 g* 2.16 J/g°C * 53 °C = 9971.21 J
Total heat = 9971.21 J + 10.905 kJ + 8479.185 J = 29355.393 J = 29.36 kJ
The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C is 29.36 kJ
Answer:
1.36 × 10³ mL of water.
Explanation:
We can utilize the dilution equation. Recall that:
Where <em>M</em> represents molarity and <em>V</em> represents volume.
Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:
Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
Convert this value to mL:
Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.
Because it’s rare to find, and has a Extraordinary color that most rocks don’t have. Please mark as brainliest
