What is the density of aluminum?

At this lower concentration, about how many extra hydrogen bonds would be needed to hold a and b together tightly enough to form

Scrat [10]

2..............................................

8 0

3 years ago

Calculate the concentration in mol L^-1 of ammonium ions in a solution prepared by mixing 360 mL of 0.250 mol L^-1 ammonium sulf

Mnenie [13.5K]

In order to determine the concentration of ammonium ions in the solution prepared by mixing solutions of ammonium sulfate, (NH4)2SO4, and ammonium nitrate, first calculate the amount of ammonium ions for each solution.

For ammonium sulfate sol'n: 0.360 L x 0.250 mol(NH4)2SO4/ L x 2 mol NH4+ /1 mol(NH4)2SO4 = 0.18 mol NH4+
For ammonium nitrate sol'n: 0.675 x 1.2 mol NH4NO3/L x 1 mol NH4+ /1 molNH4NO3 = 0.81 mol NH4+

Thus, the amount of NH4+ ions is (0.18 + 0.81) mol or 0.99 mol NH4+. To get the concentration, multiply this to the volume of solution which is assumed to be additive, such that:

M NH4+ in sol’n = 0.99 mol NH4+/1.035 L = 0.9565 mol NH4+/ L sol’n

8 0

3 years ago

2 unicellular no autotrophic It uses a flagellum. A. Kingdom animalia B. kingdom fungi B. kingdom plantae D. kingdom protista

Pachacha [2.7K]

I know the answer to the question

8 0

3 years ago

4. How can acids and bases behave in terms of electrolytes/reactivity?

icang [17]

Answer:

Strong acids are strong electrolytes. Such acids ionize completely.

Weak acids and bases are weak electrolytes. Such bases and acids ionize only to a small extent.

Explanation:

Electrolytes are the substances that produce ions when they are dissolved in water. Electrolytes can be categorized as bases, acids, and salts.

Strong acids are strong electrolytes. Such acids ionize completely.

Weak acids and bases are weak electrolytes. Such bases and acids ionize only to a small extent.

7 0

3 years ago

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .

Or, 5.0× $10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0× $10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00× $10^{-13}$ ) $^{1/2}$ = 7.071× $10^{-7}$ g/mL.

Thus the concentration of Br $^{-1}$ or HBr is 7.071× $10^{-7}$ g/mL.

4 0

4 years ago