Answer:
A. weak acid and its conjugate base
Explanation:
A buffer solution can be made with a weak acid and conjugate base or a weak base and conjugate acid.
This may help you:
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Buffers/Introduction_to_Buffers
Answer:
C-Its product is heavier than each of its reactants.
Explanation:
Correct equation:
¹⁴₇N + ¹₁H → ¹⁵₈O
In the reaction above, we can conclude within the given conditions of the reaction that the product formed is heavier than the reactants.
The product is oxygen with a mass number of 15 as shown by the superscript preceeding the symbol of the atom.
The reactants are:
Nitrogen, N with a mass number of 14
Hydrogen, H with a mass number of 1
The mass number is a true reflection of the mass of an atom. It clearly shows the mass of the nucleons which are the most massive particles that makes up an atom. The nucleons are protons and neutrons that makes up the tiny nucleus of the atom.
Oxygen here has more nucleons that each of Nitrogen and Hydrogen.
Answer:
A. 30cm³
Explanation:
Based on the chemical reaction:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
<em>1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂</em>
<em />
To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:
<em>Moles CaCO₃ -Molar mass: 100.09g/mol-</em>
1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles
<em>Moles HCl:</em>
50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles
For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:
2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.
The moles produced of CO₂ are:
2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂
Using PV = nRT
<em>Where P is pressure = 1atm assuming STP</em>
<em>V volume in L</em>
<em>n moles = 1.25x10⁻³ moles CO₂</em>
<em>R gas constant = 0.082atmL/molK</em>
<em>T = 273.15K at STP</em>
<em />
V = nRT / P
1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V
0.028L = V
28cm³ = V
As 28cm³ ≈ 30cm³
Right option is:
<h3>A. 30cm³</h3>
Answer:
c. -1020.9 kJ
Explanation:
4Fe (s) + 3 O₂ (g) --> 2 Fe₂O₃(s) ΔH = -826.0 kJ/mol.
atomic weight of iron = 56
69.03 g = 69.03 / 56
= 1.23268 moles
Heat released by 1.23268 moles
= 1.23268 x 826.0
= -1020.9 kJ .
