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Cloud [144]
4 years ago
7

Can you make a love potion

Chemistry
2 answers:
OlgaM077 [116]4 years ago
8 0
I am pretty sure you can't, but hey everything is possible! :)
yaroslaw [1]4 years ago
3 0
No unless u mix science potions together to make a love potion
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how many litres of 0.500M HCL(aq) are needed to react completely with 0.100 mol of Pb(NO3)2, forming a precipitate of PbCl2(s)..
katrin2010 [14]
The volume that is needed to react completely with 0.100 mol of lead nitrate is 0.400 L. It is because 2 moles of HCl are required for every mole of lead nitrate.
2 x 0.100 mole Pb(NO₃) = 0.200 mole HCl 
0.200 mol HCl = L x 0.500 M 
8 0
3 years ago
Read 2 more answers
What happened to the limewater in<br> the experiment? What does this<br> prove?
tensa zangetsu [6.8K]

Explanation:

In the reaction above, we clearly see that a chemical reaction took place but what kind of reaction you might ask?

Looks like it is Thermal decomposition of copper carbonate because decomposition reaction takes place due to the added heat. Decomposition in simple terms in chemical breakdown and here this result is obtained by adding heat.

1) What happened to the lime water?

Although not pictured above, but my assumption is that lime water turned milky or turbid because when CO2 comes in presence of limewater they react to form a percipitate of Calcium carbonate which is the milky color that you get.

2) What does this prove?

- My understanding would be that it proves that CO2 was formed, and that most metal carbonates undergo thermal decomposition into metal oxide and carbon dioxide, and also that a reaction took place since new products were made.

3 0
3 years ago
Balance the following redox reaction in acidic solution. Zn(s)+MnO−4(aq)→ Zn+2(aq)+Mn+2(aq)
devlian [24]

Answer:

2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺

Explanation:

To balance a redox reaction in an acidic medium, we simply follow some rules:

  1. Split the reaction into an oxidation and reduction half.
  2. By inspecting, balance the half equations with respect to the charges and atoms.
  3. In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
  4. Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
  5. Multiply both equations with appropriate factors to balance the electrons in the two half equations.
  6. Add up the balanced half equations and cancel out any specie that occur on both sides.
  7. Check to see if the charge and atoms are balanced.

Solution

                            Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺

The half equations:

                      Zn → Zn²⁺                          Oxidation half

                      MnO₄⁻ → Mn²⁺                  Reduction half

Balancing of atoms(in acidic medium)

                     Zn → Zn²⁺

                    MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balancing of charge

                   Zn → Zn²⁺ + 2e⁻

                    MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O

Balancing of electrons

         Multiply the oxidation half by 5 and reduction half by 2:

                          5Zn → 5Zn²⁺ + 10e⁻

                        2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O

Adding up the two equations gives:

              5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O

The net equation gives:

         5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O

8 0
3 years ago
For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided
SVETLANKA909090 [29]

Answer:

\frac{1}{24}

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = \frac{1}{6}

So, fraction of the whole athletic field reserved for each fifth class = \frac{1}{4}(\frac{1}{6})=\frac{1}{24}

3 0
3 years ago
In these solutions, identify the solute and the
myrzilka [38]

Answer:

1. solute

2. solvent

3.solvent

Explanation:

4 0
3 years ago
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