Distilled water is _____. an acid a base a neutral

The number of water molecules per ionic molecule in a hydrate is called the ___________. A. electronegativity B. anhydrous numbe

arlik [135]

Answer:

C. hydration number

Explanation:

When we dissolve an ionic compound (a charged species) the charges can interact with the water molecule. In the case of cations (positive charges) the negative dipole of water (generated in the oxygen) will interact with the positive charge at the same time the anions (negative charges) the positive dipole of water (generated in the hydrogen).

The amount of water molecules that can interact with a single ion (cation or anion) is called hydration number. In the example, we have a hydration number of "4" for the sodium cation.

I hope it helps!

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3 years ago

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

Colt1911 [192]

Answer:

$M_{Na^+}=1.36M$

$M_{Br^-}=1.58M$

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

$n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+$

Once we've got the moles we compute the final volume via:

$V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L$

Thus, the molarity of the sodium atoms turn out into:

$M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M$

Now, we perform the same procedure but now for the bromide ions:

$n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-$