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3 years ago

Can 1750 mL of water dissolve 4.6 moles of Copper Sulfate CuSO4? _________ Why? / Why not?

Chemistry

1 answer:

Wewaii [24]3 years ago

8 0

Answer:

<u>No, you cannot dissolve 4.6 moles of copper sulfate, CuSO₄, in 1750mL of water.</u>

Explanation:

This question is part of a Post-Lab exercise sheet.

Such sheet include the saturation concentrations for several salts.

The saturation concentration of Copper Sulfate, CuSO₄, indicated in the table is 1.380M.

That means that 1.380 moles of copper sulfate is the maximum amount that can be dissolved in one liter of solution.

Find the molar concentration for 4.6 moles of copper sulfate in 1,750 mL of water.

You need to assume that the volume of water (1750mL) is the volume of the solution. This is, that the 4.6 moles of copper sulfate have a negligible volume.

<u>1. Volume in liters:</u>

V = 1,750 mL × 1 liter / 1,000 mL = 1.75 liter

<u />

<u>2. Molar concentration, molarity, M:</u>

M = number of moles of solute / volume of solution in liters

M = 4.6 moles / 1.75 liter = 2.6 M

Since the solution is saturated at 1.380M, you cannot reach the 2.6M concentration, meaning that you cannot dissolve 4.6 moles of copper sulfate, CuSO₄ in 1750mL of water.

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3 years ago

An oxidation-reduction reaction that is also a synthesis reaction

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The example of a chemical combination that can also be said to be a synthesis reaction is; $2Mg(s) +O_{2} (g) -- > 2 MgO(s)$

<h3>What is a synthesis reaction?</h3>

We know that the meaning of the synthesis reaction is quite central as we attempt to answer the question that is lying right before us here. If we are talking about the synthesis reaction then our mind must have to go to the kind of reaction in which there is the combination of two or more so as to produce a single compound.

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11 months ago

A 75 g piece of gold (Au) at 1000 K is dropped into 200 g of H2O at 300K in an insulatedcontainer at 1 bar. Calculate the temper

Arturiano [62]

Answer:

the final temperature is T final = 308 K

Explanation:

since all heat released by gold is absorbed by water

Q gold + Q water = Q surroundings =0 (insulated)

Assuming first that no evaporation of water occurs , and denoting g as gold and w as water , then

Q gold = m g*cp g* ( T final - T initial g)

Q gold = m w*cp w* ( T final - T initial w)

where

m= mass

cp = specific heat capacity

T final = final temperature

T initial g and T initial w = initial temperature of gold and water respectively

thus

Q gold + Q water = 0

m g*cp g* ( T final - T initial g) + m w*cp w* ( T final - T initial w) =0

m g*cp g* T final + m w*cp w* T final = m g*cp g* T initial g+ m w*cp w* T initial w

T final = (m g*cp g* T initial g+ m w*cp w* T initial w)/(m g*cp g+ m w*cp w)

replacing values and assuming cp w = 1 cal/gK = 4.816 J/gK and cp g = 0.129 J/gK (from tables), then

T final = (75 g*0.129 J/gK* 1000 K + 200 g * 4.816 J/gK * 300 K )/(75 g*0.129 J/gK*+ 200 g * 4.816 J/gK ) = 308 K

T final = 308 K

since T boiling water = 373 K and T final = 308 K , we confirm that water does not evaporate

therefore the final temperature is T final = 308 K

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3 years ago

Nitrogen dioxide gas is dark brown in color and remains in equilibrium with dinitrogen tetroxide gas, which is colorless.

sukhopar [10]

Answer:

This reaction is exothermic because the system shifted to the left on heating.

Explanation:

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Reactant => NO₂ (dark brown in color)

Product => N₂O₄ (colorless)

From the question given above, we were told that when the reaction at equilibrium was moved from room temperature to a higher temperature, the mixture turned dark brown in color.

This simply means that the reaction does not like heat. Hence the reaction is exothermic reaction.

Also, we can see that when the temperature was increased, the reaction turned dark brown in color indicating that the increase in the temperature favors the backward reaction (i.e the equilibrium shift to the left) as NO₂ which is the reactant is dark brown in color. This again indicates that the reaction is exothermic because an increase in the temperature of an exothermic reaction will shift the equilibrium position to the left.

Therefore, we can conclude that:

The reaction is exothermic because the system shifted to the left on heating.

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