Answer:
Molecules that will have dipole-dipole forces with like molecules include the water (H2O) molecule. Another example is the Hydrogen Chloride (HCl) molecule.
Explanation:
Intermolecular forces are forces of attraction or repulsion that exist between particles (ions, atoms, or molecules) that are close/in nearby proximity to each other. Usually, intermolecular forces are not as strong as intramolecular forces which create covalent or ionic bonds between the atoms that exist within molecules. Dipole-dipole interactions occur whenever the partial charges that exist within one molecule are attracted to the opposite partial charges that exist within another different molecule that is nearby and similar in composition: the positive end/charges of one molecule are attracted to the negative end/charges of another similar molecule.
An example of molecules that exhibit dipole-dipole interaction is the water (H2O) molecule. Another molecule which exhibits dipole–dipole interaction is the Hydrogen Chloride (HCl) molecule, whereby the positive end of one HCl molecule usually attracts the negative end of another HCl molecule.
Subtract the number of protons from the mass number.
Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.
Answer:
1 and 2 only
Explanation: These will make a bad impact on water.
The molarity of a 27%(v/v) aqueous ethanol solution is 4.63 M
calculation
convert 27%(v/v) to fraction = 27ml/100 ml
use density to convert 27ml to grams = 27 ml x0.79 g/ml = 21.33 g
find the number of moles of C2H6O
moles = mass/molar mass of C2H6O(46.07 g/mol)
moles is therefore = 21.33 g /46.07 g/mol =0.463 moles
find the molarity = moles /volume in liters
volume in liter = 100 ml/1000 = 0.1 L
molarity is therefore = 0.463 mole/0.1 L = 4.63 M
