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Leni [432]
3 years ago
12

Balancing the reaction by oxidation number method k2cr2o7+sncl2+hcl​

Chemistry
1 answer:
Mumz [18]3 years ago
7 0

Answer:

K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)

Explanation:

The products of this reaction are given by:

K_2Cr_2O_7 (aq) + SnCl_2 (aq) + HCl (aq)\rightarrow KCl (aq) + SnCl_4 (aq) + CrCl_3 (aq) + H_2O (l)

Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

Cr_2O_7^{2-} (aq)\rightarrow Cr^{3+} (aq)

The following steps should be taken:

  • balance the main element, chromium: multiply the right side by 2 to get 2 chromium species on both side:

Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)

  • balance oxygen atoms by adding 7 water molecules on the right:

Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

  • balance the hydrogen atoms by adding 14 protons on the left:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

  • balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

Similarly, tin(II) cation becomes tin(IV) cation:

Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-

Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-

Add them together:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 3 Sn^{2+} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l) + 3 Sn^{4+} (aq)

Adding the ions spectators:

K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)

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1/2 * m * v ^ 2 = mgh
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C. the square root of 2gh

Explanation:

The square root of 2gh is equivalent to Torricelli's law in the equation.

  given expression:

         \frac{1}{2} m v² = mgh

 to find v, we make it the subject of the expression:

    *multiply both sides of the equation by 2:

             

                     2 x( \frac{1}{2} m v² )= 2 (mgh)

                                      mv²  = 2mgh

    *cancel the mass, m appearing on both sides:

                                    v²  = 2gh

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4 0
2 years ago
234+34.1 add and round to 2 significant figures
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3 years ago
Describe the properties of alkali metals. Based on their electronic arrangement, explain whether they exist alone in nature.
bearhunter [10]

Answer:

- They are highly reactive metals

- They have low electro negativity

- They have low ionization energy

- They don't exist alone in nature

- They have low densities

Explanation:

Alkali metals are the elements in group 1 of the periodic table. They include Sodium, Lithium, Potassium e.t.c.

Due to the fact they have one atom in their outermost shell, they are very unstable because they easily react with other elements and are therefore don't exist alone in nature but combined with other elements for this same reason.

Since alkali metals don't easily attract other elements due to it's lone pair in the outer most shell, it can be said to have low electro negativity.

Also, they don't need energy to discharge their electrons since they are highly reactive due to their lone pair in the outermost shell and so we say they have low ionization energy.

Due to this reason, they also have low densities.

7 0
2 years ago
Calculate the mass fraction of sodium chloride in the solution if 20 g of it is dissolved in 300 ml of water.
MA_775_DIABLO [31]

The mass fraction of sodium chloride is 0.0625

<h3>What is the mass fraction of sodium chloride in the solution?</h3>

The mass fraction of sodium chloride is the ratio of the mass of sodium chloride to the total mass of the solution.

The mass fraction of sodium chloride is determined as follows;

mass of sodium chloride = 20 g

  • mass of water = volume * density

density of water = 1 g/mL

volume of water = 300 mL

mass of water = 300 mL * 1 g/mL

mass of water = 300 g

total mass of solution = 20 + 300 = 320 g

mass fraction of sodium chloride = 20/320

mass fraction of sodium chloride = 0.0625

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