Answer : The amount of codeine in one teaspoonful is, 30 mg
Explanation : Given,
Amount of codeine per fluid ounce = 60 mg
Now we have to determine the amount of codeine present in one teaspoonful.
As we know that:
1 teaspoonful = 0.5 fluid ounce
As, the amount of codeine per fluid ounce = 60 mg
So, the amount of codeine 0.5 fluid ounce =
Thus, the amount of codeine in one teaspoonful is, 30 mg
Answer:
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Answer:
The product is 1-methylcyclopentanol
Explanation:
The acid-catalyzed hydration of alkenes involves the addition of a water molecule to a C=C double bond.
H₂O + C=C ⟶ H-C-C-OH
An H atom adds to one of the C atoms, and an OH group adds to the other
This reaction follows Markovnikov’s rule — the H adds to the C atom that has more hydrogen atoms, and the OH adds to the more substituted carbon.
The steps of the mechanism are:
- The aqueous sulfuric acid generates hydronium ions
- The nucleophilic π electrons attack an H atom on the hydronium ion, forming a carbocation on the more substituted C atom.
- The lone pair electrons on a water molecule attack the carbocation , forming an oxonium ion.
- Another water molecule removes the extra proton.
The product is the alcohol with the OH group on the more substituted carbon — 1-methylcyclopentanol.
[NH₄⁺] = 1.64 × 10⁻³ M
Explanation:
We have the following chemical equilibrium:
NH₃ + H₂O ⇔ NH₄⁺ + H O⁻
Kb is defined:
Kb = ( [NH₄⁺] × [H O⁻] ) / [NH₃]
and if we rearrange the therms:
Kb × [NH₃] = [NH₄⁺] × [H O⁻]
[NH₄⁺] × [H O⁻] = Kb × [NH₃]
The concentration of ammonium ion [NH₄⁺] is equal to the concentration of hydroxide ion [H O⁻]. Taking this in account we get:
[NH₄⁺]² = Kb × [NH₃]
[NH₄⁺]² = 1.80 × 10⁻⁵ × 0.150
[NH₄⁺] = √ (1.80 × 10⁻⁵ × 0.15)
[NH₄⁺] = 1.64 × 10⁻³ M
Learn more about:
problems with acid and bases
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Louis Pasteur's experiment was A) <em>controlled</em>.
A scientist must change <em>only one variable at a time</em> to see if it is the one causing the observed change.
The scientist must keep all the other variables constant (i.e. <em>control </em>the variables)