Answer:
the golf ball would move faster, but the bowling ball would have more kinetic energy
Explanation:
Answer:
7.5×10²³ ions
Explanation:
Given data:
Mass of magnesium sulfate = 75.0 g
Number of ions = ?
Solution:
Number of moles of magnesium sulfate:
Number of moles = mass/molar mass
Number of moles = 75.0 g/ 120.36 g/mol
Number of moles = 0.623 mol
Magnesium sulfate formula: MgSO₄
Number of moles of ions in 1 mole of MgSO₄ = 2 mol (Mg²⁺ and SO₄²⁻)
Number of moles of ion in 0.623 moles of MgSO₄:
2 mole ×0.623 = 1.246 mol
Number of ions:
1 mole contain 6.022×10²³ ions
1.246 mol × 6.022×10²³ ions / 1mol
7.5×10²³ ions
Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
We have that The the mass change when the copper coin was made to look silver is an increase in mass
Correct option C
It increased.
It is important to note that the copper coin after its cutting into shape will have a specific mass or weight and the silver coating solution will also have a net value of mass or weight
Therefore
The the mass change when the copper coin was made to look silver is an increase in mass
Correct option C
It increased.
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There are 1000 mililiters in a liter, so 1000 ml for every liter, you have 5 liters, so:
5L*1000 = 5000 mL
