There is no reaction.
<em>Molecular equation
:</em>
K₂CO₃(aq) + 2NH₄Cl(aq) ⟶ 2KCl(aq) + (NH₄)₂CO₃(aq)
<em>Ionic equation
:</em>
2K⁺(aq) + CO₃²⁻(aq) + 2NH₄⁺(aq) +2Cl⁻(aq) ⟶ 2K⁺(aq) + 2Cl⁻(aq) + 2NH₄⁺(aq) + CO₃²⁻(aq)
<em>Net ionic equation
:</em>
Cancel all ions that appear on both sides of the reaction arrow (underlined).
<u>2K⁺(aq)</u> + <u>CO₃²⁻(aq)</u> + <u>2NH₄⁺(aq</u>) +<u>2Cl⁻(aq)</u> ⟶ <u>2K⁺(aq)</u> + <u>2Cl⁻(aq</u>) + <u>2NH₄⁺(aq)</u> + <u>CO₃²⁻(aq)</u>
<em>All ions cancel</em>. There is no net ionic equation.
Answer:
The carbocation intermediate reacts with a nucleophile to form the addition product.
Explanation:
The reaction of benzene with an electrophile is an electrophillic substitution reaction. Here the electrophile replaces hydrogen. There is no formation of carbocation as intermediate in the reaction. Infact there is transition state where the electorphile attacks on benzene ring and at the same time the hydrogen gets removed from the benzene. So a transition carbocation is formed.
The general mechanism is shown in the figure.
i) Attack of the electrophile on the benzene (which is the nucleophile)
ii) The carbocation intermediate loses a proton from the carbon bonded to the electrophile.
iii) the carbocation formation is the rate determining step.
iv) There is no formation of addition product.
Thus the wrong statement is
The carbocation intermediate reacts with a nucleophile to form the addition product.
Answer:
The concentration is 0.036 mg/mL
Explanation:
Concentration = 0.2 mM = 0.2/1000 = 2×10^-4 M = 2×10^-4 mol/L × 180,000 mg/1 mol × 1 L/1000 mL = 0.036 mg/mL
Scientists use carbon-14 to date ancient fossils.
Plants and animals absorb carbon-14 produced by cosmic rays. The ratio of C-14 to C-12 is constant when they are alive.
When they die, the C-14 decays to C-12, and the ratio changes.
Carbon-14 has a half-life of 5730 years. Thus, the object loses half its C-14 every 5730 years.
Scientists can use the C-14 to C-12 ratio to date fossils up to 70 000 years old.
Burning is a chemical property so (c) for #1
Sodium has one valence electron in its 3rd orbit
