Freezing point depression depends of the number of particles of the solute in the solution.
1)Pure water have highest freezing point. All other solutions with given solutes will have lower temperatures.
2) The more particles of the solute in the solution the lower freezing point is going to be.
<span>b. 1.0 m NaCl ( dissociates and give 2 mol ions (1 mol Na⁺ and 1 mol Cl⁻))
c. 1.0 m K3PO4 (</span>dissociates and give 4 mol ions (3 mol K⁺ and 1 mol PO4³⁻)<span>
d. 1.0 m CaCl2 (</span>dissociates and give 3 mol ions (1 mol Ca²⁺ and 2 mol Cl⁻))<span>
e. 1.0 m glucose (c6h12o6) (glucose does not dissociate, and solution have
1 mole of particles of the solute(glucose))
The largest number of particles has </span>1.0 m K3PO4 solution, and it is has lowest freezing point . Answer is C.
Answer:
See explanation
Explanation:
The reaction that we are considering here is quite a knotty reaction. It is difficult to decide if the mechanism is actually E1 or E2 since both are equally probable based on the mass of scientific evidence regarding this reaction. However, we can easily assume that the methylenecyclohexane was formed by an E1 mechanism.
Looking at the products, one could convincingly assert that the reaction leading to the formation of the two main products proceeds via an E1 mechanism with the formation of a carbocation intermediate as has been shown in mechanism attached to this answer. Possible rearrangement of the carbocation yields the 3-methylcyclohexene product.
Answer:
This would be the hydride ion where H gains an electron from a metal such as Na
Explanation:
Answer:

Explanation:
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In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

We plug in the mass of water, temperature change and specific heat to obtain:

Now, this enthalpy of reaction corresponds to the combustion of propyne:

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

Now, we solve for the enthalpy of formation of C3H4 as shown below:

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

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