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denis23 [38]
3 years ago
15

A salt is made by adding an excess of an insoluble metal oxide to an acid.

Chemistry
1 answer:
Blizzard [7]3 years ago
3 0

Answer:

Filtration

Explanation:

Filtration is used to separate solid substances from liquids or large molecules from small molecules. Since, Acid is a liquid and metal oxide is solid so Filtration can be used to separate these two.

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Which of the following equations is balanced?
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Where are the equations ..

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What led Marie Curie to draw the following conclusions? (b) A highly radioactive element, aside from uranium, occurs in pitchble
AVprozaik [17]

Making repeated separations of the various substances in the pitchblende, Marie and Pierre used the Curie electrometer to identify the most radioactive fractions. They thus discovered that two fractions, one containing mostly bismuth and the other containing mostly barium, were strongly radioactive.

<h3>What was surprising about pitchblende?</h3>

Since it was no longer appropriate to call them “uranic rays,” Marie proposed a new name: “radioactivity.”

Even more surprising, Marie next found that a uranium ore called pitchblende contained two powerfully radioactive new elements: polonium, which she named for her native Poland, and radium.

<h3>Why is radium more radioactive than uranium?</h3>

It is 2.7 million times more radioactive than the same molar amount of natural uranium (mostly uranium-238), due to its proportionally shorter half-life.

Learn more about highly radioactive elements here:

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6 0
2 years ago
In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an ex
Kay [80]

The question is incomplete. The complete question is :

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.

Solution :

Consider the equation of the reaction between NaOH and $HCl$

  NaOH (aq) + HCl (aq) → NaCl(aq) + $H_2O (l)$

The above equation tells us that $1 \text{mole}$ of $NaOH$ reacts with $1 \text{mole}$ of $HCl$.

So at the equivalence point, the moles of NaOH added = moles of $HCl$present.

If the volume of the $HCl$ taken = $V_1$ mL and the conc. of $HCl$ = $M_1$  mole/L

The volume of NaOH added up to the color change = $V_2 \text{  and conc of NaOH = M}_2$ mole/L

Moles of $HCl$ taken = $V_1 \ mL \times M_1 \ mol/100 \ mL = V_2M_2 \times 10^{-3}$  moles.

The color change will occur when the moles of NaOH added is equal to the moles of $HCl$ taken.

Thus when $V_1 M_1 \times 10^{-3} = V_2M_2 \times 10^{-3}$

or   when    $V_1M_1 = V_2M_2$

or $V_2=\frac{V_1M_1}{M_2}$  mL of NaOH added, we observe the color change.

Where $V_1, M_1$ are the volume and molarity of the $HCl$ taken.

$M_2$ is the molarity of NaOH added.

When both the NaOH and $HCl$ are of the same concentrations, i.e. if $M_1=M_2$, then $V_2=V_1$

Or the 40 mL of $HCl$ will need 40 mL of NaOH for a color change and

30 mL of $HCl$ would need 30 mL of NaOH for the color change (provided the concentration $M_1=M_2$)

7 0
3 years ago
One solution turns blue. A possible hydrogen ion
aalyn [17]

Answer:

1x10^-8 M

Explanation:

Since the solution turns blue, it mean the solution is a base.

Now, to know which option is correct, we need to determine the pH of each solution. This is illustrated below:

1. Concentration of Hydrogen ion, [H+] = 1x10^-2 M

pH =..?

pH = - log [H+]

pH = - log 1x10^-2

pH = 2

2. Concentration of Hydrogen ion, [H+] = 5x10-2 M

pH =..?

pH = - log [H+]

pH = - log 5x10^-2

pH = 1.3

3. Concentration of Hydrogen ion, [H+] = 5x10 M

pH =..?

pH = - log [H+]

pH = - log 5x10

pH = - 1.7

4. Concentration of Hydrogen ion, [H+] = 1x10-8 M

pH =..?

pH = - log [H+]

pH = - log 1x10^-8

pH = 8

A pH reading shows if the solution is acidic or basic. A pH reading between 0 and 6 indicates an acidic solution, a pH reading of 7 indicates a neutral solution while a pH reading between 8 and 14 indicates a basic solution.

From the above calculations, the pH reading indicates a basic solution when the hydrogen ion concentration was 1x10^-8 M.

5 0
3 years ago
Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the conce
levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca^{2+}(aq) and 0.0390 M Ag^{+}(aq). What will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate? What percentage of the Ca^{2+}(aq) can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

      Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}

[Ag^{+}] = 0.0390 M

When Ag_{2}SO_{4} precipitates then expression for K_{sp} will be as follows.

         K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]

        1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]

       [SO^{2-}_{4}] = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

         CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}

      K_{sp} = [Ca^{2+}][SO^{2-}_{4}]

     4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788

           [Ca^{2+}] = 0.00625 M

Now, we will calculate the percentage of Ca^{2+} remaining in the solution as follows.

               \frac{0.00625}{0.05} \times 100

                 = 12.5%

And, the percentage of Ca^{2+} that can be separated is as follows.

                     100 - 12.5

                     = 87.5%

Thus, we can conclude that 87.5% will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate.

4 0
3 years ago
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