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Marizza181 [45]
1 year ago
11

22.6 grams of mercury ii nitrate, hg(no3)2 reacts with an excess of potassium, k. how many grams of mercury is formed? the other

product is potassium nitrate, kno3.
Chemistry
1 answer:
pentagon [3]1 year ago
5 0

Mass = 14.0 g of mercury is formed, 22.6 grams of mercury nitrate, hg(no3)2 reacts with an excess of potassium, k.

An inorganic substance with the formula Hg(NO3)2xH2O is mercury nitrate. Hot, concentrated nitric acid is used to process mercury to create it. By using X-ray crystallography, neither anhydrous nor monohydrate have been verified. The drug might have an impact on the kidneys, peripheral nervous system, and central nervous system. Ataxia, tremors. Given data: Mass of mercury nitrate = 22.17 g, Mass of mercury formed = ?

Solution:

Chemical equation:

Hg(NO₃)₂ + 2K    →    2KNO₃ + Hg

Number of moles of mercury nitrate:, Number of moles = mass/molar mass, Number of moles = 22.17 g / 324.6 g/mol, Number of moles = 0.07 mol, Now we will compare the moles of Hg(NO₃)₂ and mercury.

             Hg(NO₃)₂       :        Hg

                 1                :          1

              0.07             :       0.07

Mass of mercury: Mass = number of moles × molar mass, Mass = 0.07 mol × 200.6 g/mol

Mass = 14.0 g

Learn more about mercury nitrate here:

brainly.com/question/4588222

#SPJ4

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Without doing any calculations, determine which of the following solutions would be most acidic.
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Answer:

In the first combination neutralization takes place to give a salt. So, solution 'a' is neutral in nature.

In the solution 'c', both salts are resulted by the combination of weak base and strong acid. The combination of these salts suppresses the acidity.

In last combination basic nature is observed due to the presence of CN⁻ ions. Thus, the solution 'd' is basic in nature.

Out of the five given solutions, 0.0100 M in HF and 0.0100 M in KBr is most acidic.  Therefore, solution 'b' is most acidic in nature.

Explanation:

5 0
3 years ago
How many moles of electrons is required to deposit 5.6g of iron from a solution of iron (2) tetraoxosulphate(6)
yuradex [85]

Answer:

0.20 mol

Explanation:

Let's consider the reduction of iron from an aqueous solution of iron (II).

Fe²⁺ + 2 e⁻ ⇒ Fe

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5.6 g × 1 mol/55.85 g = 0.10 mol

2 moles of electrons are required to deposit 1 mole of Fe. The moles of electrons required to deposit 0.10 moles of Fe are

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7 0
3 years ago
12. Describe the results of a chemical change. List four<br> indicators of chemical change.
marshall27 [118]

Answer:

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8 0
3 years ago
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tino4ka555 [31]

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3 years ago
Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of
Brut [27]

<u>Answer:</u> The smallest temperature change is shown by water.

<u>Explanation:</u>

To calculate the heat absorbed or released, we use the equation:

q=mC\times \Delta T      ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

\Delta T = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

  • <u>Option a:</u>  50.0 g Fe, C_{Fe}=0.449J/g^oC

We are given:

m=50.0g\\C_{Fe}=0.449J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC

Change in temperature = 8.99°C

  • <u>Option b:</u>  50.0 g water, C_{water}=4.18J/g^oC

We are given:

m=50.0g\\C_{water}=4.18J/g^oC

Putting values in equation 1, we get:

200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC

Change in temperature = 0.96°C

  • <u>Option b:</u>  25.0 g Pb, C_{Pb}=0.128J/g^oC

We are given:

m=50.0g\\C_{Pb}=0.128J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC

Change in temperature = 62.5°C

  • <u>Option d:</u>  25.0 g Ag, C_{Ag}=0.235J/g^oC

We are given:

m=25.0g\\C_{Ag}=0.235J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC

Change in temperature = 34.04°C

  • <u>Option e:</u>  25.0 g granite, C_{granite}=0.79J/g^oC

We are given:

m=25.0g\\C_{Fe}=0.79J/g^oC

Putting values in equation 1, we get:

200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

5 0
3 years ago
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