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QveST [7]
3 years ago
8

HELP ME PLEASE!! If given a graduated cylinder, a beaker with alcohol and a scale, explain the steps to find the mass of 40 mL o

f alcohol.
Chemistry
1 answer:
Luda [366]3 years ago
6 0
<h3>Further explanation </h3>

In general, some people equate mass and weight.

Mass is one of the principal quantities, which is related to the matter , whereas weight is a force that leads to the center of the earth (Earth's gravitational force)

Steps that can be taken  to find the mass :

  • 1. Weigh graduated cylinder (empty)
  • 2. Pour the alcohol in the beaker into the graduated cylinder to the level of 40 ml
  • 3. Weigh again graduated cylinder + poured alcohol
  • 4. mass of alcohol 40 ml = mass in 3rd step - mass in the first step

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Write a net ionic equation for the reaction that occurs when excess hydrochloric acid (aq) and potassium sulfite (aq) are combin
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<u>Answer:</u> The net ionic equation for the given reaction is 2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow H_2O(l)+SO_2(g)

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are the ions which do not get involved in a chemical equation. It is also defined as the ions that are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of hydrochloric acid and potassium sulfite is given as:

2HCl(aq.)+K_2SO_3(aq.)\rightarrow 2KCl(aq.)+SO_2(g)+H_2O(l)

Ionic form of the above equation follows:

2H^+(aq.)+2Cl^-(aq.)+2K^+(aq.)+SO_3^{2-}(aq.)\rightarrow 2K^+(aq.)+2Cl^-(aq.)+SO_2(g)+H_2O(l)

As, potassium and chloride ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow SO_2(g)+H_2O(l)

Hence, the net ionic equation for the given reaction is written above.

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In the reaction 2N2O <-----> O2 + 2N2, adding N2O to the mixture favors the forward reaction and more products are formed, equilibrium shifts to the right.

Also, removing the product formed favors the forward reaction as more products will be formed in order to achieve equilibrium. The equilibrium therefore shifts to the right.

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