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Jobisdone [24]
3 years ago
13

How many grams of Ag2CO3 are required to reacr with 28.5 mL of 1.00 M NaOH solution?

Chemistry
1 answer:
schepotkina [342]3 years ago
8 0
The reaction between Ag2CO3 and NaOH is shown by the equation below 
Ag2CO3 + NaOH = Ag2O + Na2CO3 +H2O
we can determine the number of mole of sodium hydroxide 
by (2.85 ml × 1) ÷ 1000 ml , since according to molarity 1 mole is contained in 100ml. 
we get 0.00285 moles of NaOH
Using the mole ratio we can get the moles of Ag2CO3
Mole ratio: Ag2NO3 : NaOH  = 1:1 
Therefore, the moles of Ag2CO3 will be 0.00285 moles
but 1 mole of silver carbonate is equivalent to 275.8 g
Thus the mass will be calculated by 0.00285 moles × 275.8g = 0.78603g
Mass of silver carbonate required will be 0.78603g

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3 years ago
The heat of vaporization of a liquid is 84.0 J/g. How many joules of heat would it take to completely vaporize 172 g of this liq
Aliun [14]

Answer:

14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.

Explanation:

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.

During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.

So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.

In this case, being:

  • Q=?
  • m= 172 g
  • L= 84 \frac{J}{g}

and replacing in the expression Q = m*L you get:

Q=172 g*84 \frac{J}{g}

Q=14,448 J

<u><em>14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.</em></u>

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