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3 years ago

Solutions are formed by____.

1 answer:

3 0

1.In order to form a solution, the solute must be surrounded, or solvated, by the solvent. Solutes successfully dissolve into solvents when solute-solvent bonds are stronger than either solute-solute bonds or solvent-solvent bonds.

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

A motorcycle travel at 6.0 m/s. After 3.0 second, the motorcycle travel at 15.0 m/s. Which of the following was the average acce

Lerok [7]

Hello!

$\large\boxed{a = 3m/s^{2}}$

Use the following equation to solve for the average acceleration of the motorcycle:

$a = \frac{v_{f}-v_{i}}{t}$

Plug in the given final, initial velocities, and the time:

$a = \frac{15-6}{3}\\\\a = \frac{9}{3}\\\\a = 3m/s^{2}$

7 0

3 years ago

Can someone please help me, I’m taking a test

lawyer [7]

Answer:

Im pretty sure its none of these

8 0

3 years ago

Write a net ionic equation for the overall reaction that occurs when aqueous solutions of nitrous acid and sodium hydroxide are

Bess [88]

Answer:

H+(aq) + OH-(aq) → H2O(l)

Explanation:

Step 1: Data given

nitrious acid = HNO3

sodium hydroxide = NaOH

Step 2: The unbalance equation

HNO3(aq) + NaOH(aq) →NaNO3(aq) + H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side (Ba^2+ and Br-), look like this:

H+(aq) + NO3-(aq) + Na+(aq) + OH-(aq) →Na+(aq) +NO3(aq) + H2O(l)

H+(aq) + OH-(aq) → H2O(l)

7 0

4 years ago

If a solution containing

tankabanditka [31]

Answer:

Hg(NO₃)₂(aq) + Na₂SO₄(aq) → 2NaNO₃(aq) + HgSO₄(s)

Moles of Hg(NO₃)₂ = 55.42 / 324.7 ==> 0.1707 moles

Moles of Na₂SO₄ = 16.642 / 142.04 ==> 0.1172 moles

Limiting reagent is Na₂SO₄ as it controls product formation

Moles of HgSO₄ formed = 0.1172 moles

= 0.1172 x 296.65

= 34.757g

Explanation:

6 0

3 years ago