Original molarity was 1.7 moles of NaCl
Final molarity was 0.36 moles of NaCl
Given Information:
Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity
Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity
1. Solve for the molarity of the original (concentrated) solution.
Molarity (M) = moles of solute (mol) / liters of solution (L)
Convert the given information to the appropriate units before plugging in and solving for molarity.
Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)
2. Solve for the molarity of the final (diluted) solution.
Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.
Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution
Molarity (M) of the final solution = 0.36 M NaCl
I hope this helped:))
(A)Nuclear change..............
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
<span>D. The screw changes the direction of a force - it converts linear force into rotational force. It also reduces the force required - the closer the threads, the smaller the input force required to get the same output force.</span>
ANSWER:
The melting and boiling points increase in order of increasing atomic number.
The size of the nucleus increases in order of increasing atomic number.
Ionization energy decreases in order of increasing atomic number.
Electronegativity decreases in order of increasing atomic number.
Electron Affinity decreases in order of increasing atomic number.
The reactivities decrease in order of increasing atomic number.
EXPLANATION:
NAME MELTING POINT BOILING POINT
Fluorine -220 -188
Chlorine -101 -35
Bromine -7.2 58.8
Iodine 114 184
Melting and Boiling points increase as shown above.
NAME COVALENT RADIUS IONIC RADIUS
Fluorine 71 133
Chlorine 99 181
Bromine 114 196
Iodine 133 220
Size increases as shown above.
NAME FIRST IONIZATION ENERGY
Fluorine 1681
Chlorine 1251
Bromine 1140
Iodine 1008
Ionization energy decreases as shown above.
NAME ELECTRONEGATIVITY
Fluorine 4
Chlorine 3
Bromine 2.8
Iodine 2.5
Electronegativity decreases as shown above.
NAME ELECTRON AFFINITY
Fluorine -328.0
Chlorine -349.0
Bromine -324.6
Iodine -295.2
Electron affinity decreases as shown above.
REACTIVITY
The reactivities of the halogens decrease. This is due to the fact that atomic radius increases in proportion with an increase of electronic energy levels. This decreases the pull for valence electrons of other atoms, minimizing reactivity.
