Answer:
See the answer and explanation below , please.
Explanation:
A conjugate base is defined as that formed after an acid donates its proton.
For each article, a continuation of the conjugate bases (highlighted in bold), for dissociation in water:
a) HF + H20 --> F- + H30+
b) H20+ H20 --> OH- + H30+
C)H2PO3- + H20--> HPO3 2- + H30+
d) HSO4- + H20 --> SO4 2- + H30+
E) HCL02 + H20 --> CLO02 - + H30+
Answer:
A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)
Explanation:
Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.
The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:
Ca(s), C(s) and O₂(g)
That means, the equation that represents standard enthalpy of CaCO₃ is:
<h3>A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)</h3><h3 />
<em>Is the equation that has ΔH° = -1207kJ/mol</em>
Answer:
CH3COOH would be more concentrated
Explanation:
The higher the concentration value, the more concentrated it is.
The relationship between concentration, moles and volume is given by the equation;
Concentration = No of moles / Volume
5.0 grams of HCOOH dissolved in 189 mL of water
Number of moles = Mass / Molar mass = 5 / 46.03 = 0.1086 mol
Concentration = 0.1086 / 0.189 = 0.5746 mol/L
1.5 moles of CH3COOH dissolved in twice as much water
Volume = 2 * 189 = 378 ml = 0.378 L
Concentration = 1.5 / 0.378 = 3.9683 mol/L
Comparing both concentration values;
CH3COOH would be more concentrated
Simply look at the periodic table and fill in what you know based on the table
The number of protons = atomic number
The number of electrons, Which is the same as the atomic number for atoms.
The number of valence electrons that is given by the group that the element is in, the top number of each column in the periodic table.
The atomic mass of K is 39
from Avogadro's law
39g of K contains 6.02x10^23 atoms
therefore if
39=6.02x19^23
X=5.11×10^22
making X the subject of the formula
X= (5.11×10^22×39)÷6.02×10^23
X= 33g
