Answer:
Explanation:
Its just dangerous stuff can go in the air and harm others
This question is incomplete, the complete question is;
Tonksite is a solid at 300.00K. At 300.00 K its enthalpy of sublimation is 66.00 kJ/mol. The sublimation pressure at 300.00 K is 5.00 × 10⁻⁴ atm
Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.
Answer: the sublimation pressure of the solid at the melting point is 0.3727 atm
Explanation:
Given that;
T1 = 300 K
T2 = 400 K
H_sub = 66 kJ/mol = 66000 J/mol
P1 = 5.00 × 10⁻⁴ atm
p2 = ?
now using the expression
log( p2 / 5.00 × 10⁻⁴ ) = (H_sub / R × 2.303 ) (( T2 - T1) / T1T2)
now we substitute of given values into the expression
log(p2/p1) = (66000 / 8.314 × 2.303 ) (( 400 - 300) / 300 × 400 )
p2 = 0.3727 atm
therefore the sublimation pressure of the solid at the melting point is 0.3727 atm
Answer:
The conversion of liquid water into gaseous water is a chemical change
Explanation:
A chemical change occurs when there is a chemical reaction, so there'll be changed in the compounds, such as forming new ones, forming its elements, or elements forming compounds.
A physical change occurs when there is a change in the state of aggregation of the compound, it means that it changes its physical state. Solid for liquid, liquid for solid, liquid for gas, gas for liquid, solid for gas, and gas for solid are the physical changes.
So the evaporation of water, or its conversion in gaseous water, is a physical change, not a chemical change.
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl
Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.
To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH
Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH
Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH
Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH
I hope this helps.
