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6.3072 g >>molarity = moles of solute/volume of solution (in litres) 0.45 m = n/0.4 l n = 0.45 m × 0.4 l = 0.18 mol you need 0.18 mol of nh_4oh molar mass of nh_4oh is 35.04 g/mol mass of solute = 0.18 cancelmol × 35.04 g/cancelmol = 6.3072 g

This is also a 1:1 ratio. Generally, oh adds hydroxide to an inorganic compound's name Moreover, element names aren't capitalized unless at the beginning of a sentence We write iron (ii) hydroxide instead of just iron hydroxide as iron takes the form of its +2 oxidation state, out of its 10 oxidation states. So this is a propanol derivative Both names seem to be unambiguous.

No, the oh group in phenol is by definition at c1 > the structure of phenol is you can draw the structure like this It is phenol because the ring carbon attached to the oh group is now c1 The oh carbon can be on any carbon atom of the ring, and the compound is still phenol, If naoh is added, then we have increased the concentration of hydroxide ions We want the standard enthalpy of formation for ca (oh)_2

Thus, our required equation is the equation where all the constituent elements combine to form the compound, i.e.

Because n aoh → n a+ + oh − and n i2+ + 2oh −→n i(oh)2 total mols of n aoh 60ml ⋅ 0.45m ol/l = 27mm ol of n aoh (dont forget the m for milli=one thousandth) this will produce 27 2 = 13.5mm ol of n i(oh)2 now you multiply this by the molecular mass number to get the weight in milligrams (divide by 1000 to get the grams) Nickel (ii)hydroxide may be formed without or (more likely. In a covalent bond between two atoms of unequal electronegativity, the more electronegative atom draws electron density towards itself This causes the δ⁺ and δ⁻ charges of the bond dipole The acid in excess is then titrated with n aoh (aq) of known concentration.we can thus get back to the concentration or molar quantity of m (oh)2.as it stands the question (and answer) are hypothetical.

Oh− (aq) + h3o+ (aq) → 2h2o(l) so you can say that when you mix these two solutions, the hydronium cations present in the hydrochloric acid solution will be the limiting reagent, i.e They will be completely consumed by the reaction. The added water to reach 100.00 ml doesn't change the mols of hcl present, but it does decrease the concentration by a factor of 100//40 = 2.5 Regardless, what matters for neutralization is what amount of naoh you add to what number of mols of hcl I got ph's of 1.36, 1.51, 1.74, 2.54 You started with 0.1100 m hcl, but it was diluted from 40 ml to 100 ml