Answer: B
Explaination: The liquid solution changed color means is turned into a different substance with different properties. It is chemical change .
<span>moles water = 850.0 g / 18.02 g/mol=47.2
moles Mg2+ = 0.400
moles Cl- = 2 x 0.400 = 0.800
moles ions = 0.400 + 0.800= 1.2
mole fraction ions = 1.2 / 1.2 + 47.2 =0.0248</span>
Answer:
A strong base, such as NaOH. The amount of OH added shouldn't exceed 0.35 mols (though i would stop at 0.30 mols)
Explanation:
a weakly basic salt can be turned into a buffer by the addition of a strong base, and a weakly acidic salt can be turned into a buffer with a strong acid
Answer:
²³⁸₉₂U has 146 neutrons
Explanation:
To know which option is correct, we shall determine the number of neutrons in each case. This can be obtained as follow:
For ²⁴₁₂Mg:
Mass number (A) = 24
Proton number (P) = 12
Neutron number (N) =?
A = P + N
24 = 12 + N
Collect like terms
N = 24 – 12
N = 12
Thus, there are 12 neutrons in ²⁴₁₂Mg
For ⁷⁵₃₃As:
Mass number (A) = 75
Proton number (P) = 33
Neutron number (N) =?
A = P + N
75 = 33 + N
Collect like terms
N = 75 – 33
N = 42
Thus, there are 42 neutrons in ⁷⁵₃₃As
For ¹⁹₉F:
Mass number (A) = 19
Proton number (P) = 9
Neutron number (N) =?
A = P + N
19 = 9 + N
Collect like terms
N = 19 – 9
N = 10
Thus, there are 10 neutrons in ¹⁹₉F.
For ²³⁸₉₂U:
Mass number (A) = 238
Proton number (P) = 92
Neutron number (N) =?
A = P + N
238 = 92 + N
Collect like terms
N = 238 – 92
N = 146
Thus, there are 146 neutrons in ²³⁸₉₂U
Considering the options and the illustration made above, Only ²³⁸₉₂U has the correct number of neutrons written along side.
Answer:
See explanation
Explanation:
The nucleophile here is CH3OH. We know that CH3OH is a good nucleophile that promotes SN2 reanction. However, (R)-6-bromo-2,6-dimethylnonane is a tertiary alkyl halide so the reaction proceeds by SN1 mechanism. This means that a racemic mixture is obtained at the end of the reaction because the attack occurs at the stereogenic carbon atom (6R) hence the product is optically inactive.
On the other hand, when (5R)-2-bromo-2,5-dimethylnonane is reacted with CH3OH, an optically active product is obtained because; though a tertiary alkyl halide and reaction occurs by SN1 mechanism, the attack does not occur at the stereogenic carbon atom (5R). Therefore, an optically active product is obtained in this case.
