Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>
Answer:
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)
Explanation:
Write the balance chemical equation ,

initial concenration of 
lets assume that degree of dissociation=
concenration of each component at equilibrium:
![[SO_2Cl_2] = 0.1-0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha)
![[SO_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2%5D%20%3D%200.1%5Calpha)
![[Cl_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BCl_2%5D%20%3D%200.1%5Calpha)


as
is very small then we can neglect 
therefore ,



Eqilibrium concenration of ![[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha%20%3D%200.1-0.1%5Ctimes%200.00173)
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)
Answer:
A, 18.4 and 34!
Explanation:
The Burette seems to have <em><u>~18.4</u></em> of the liquid? substance in it, and the Measuring Cylinder seems to have <em><u>~34</u></em> of the liquid? substance in it. <em><u>So, it is A!</u></em> :]
Hope this helps!!
Answer:
See explanation below
Explanation:
The first two pictures show the reagents used in these reactions a) and b). As it was stated, An E2 reaction proceeds with an antiperiplanar stereochemistry, so in the case of reaction a) it fill form a product with the groups in opposite directions. In other words, a Trans product.
In the case of reaction b) we have the same reaction, with the difference that we have changed the CH3 and phenyl group of positions. This will cause that the reaction will proceed the same but the stereochemistry of the final product will be changed too. In this case, and according to the picture 3 attached, we can see that the product formed is a cis product. So we can conclude that the relation of product a) and b) is that they are isomers, the trans and cis isomers respectively. See picture below for mechanism and products