1) Chemical reaction: HI(aq) → H⁺(aq) + I⁻(aq).
It gives an iodide anion.
2) Chemical reaction: H₂O → H⁺(aq) + OH⁻(aq).
It gives a hydroxide anion.
3) Chemical reaction: NH₄⁺(aq) → H⁺(aq) + NH₃(aq).
It gives ammonia.
4) Chemical reaction: HNO₃(aq) → H⁺(aq) + NO₃⁻(aq).
It gives nitrate anion.
I dont know kid, gotta ask someone else.
Answer:
Oxidation: a type of chemical reaction where one or more electrons are lost.
Oxidation State / Number: a number assigned to an atom describing its degree of oxidation, meaning how many electrons it has gained or lost.
Reduction: a type of chemical reaction where one or more electrons are gained.
Oxidation-Reduction Reaction: a chemical reaction where oxidation and reduction occurs simultaneously
Explanation:
Reduction always occurs at cathode
Oxidation always occurs in anode
These two process occurs in same way independent of nature of cell whether voltaic or electrolytic.
Answer:
4Ba(CO3) -> 4BaO2 + 2CO2
Explanation:
I looked at the oxygens to balance this. Ba(CO3) normally has 3 oxygens. BaO2 and CO2 have 4 oxygens total. The common multiple of 3 & 4 is 12. So there should be 12 oxygens on both sides. Then I just found the coefficients that would give 12 oxygens on both sides and can balance the rest of the atoms.
Answer:
0.0055 mol of N2O5 will remay after 7 min.
Explanation:
The reaction follows a first-order.
Let the concentration of N2O5 after 7 min be y
Rate = Ky = change in concentration of N2O5/time
K is rate constant = 6.82×10^-3 s^-1
Initial concentration of N2O5 = number of moles/volume = 2.1×10^-2/1.8 = 0.0117 M
Change in concentration = 0.0117 - y
Time = 7 min = 7×60 = 420 s
6.82×10^-3y = 0.0117 - y/420
0.0117 - y = 420×6.82×10^-3y
0.0117 - y = 2.8644y
0.0117 = 2.8644y + y
0.0117 = 3.8644y
y = 0.0117/3.8644 = 0.00303 M
Number of moles of N2O5 left = y × volume = 0.00303 × 1.8 = 0.0055 mol (to 2 significant digits)
