Why does magnesium nitrate not react with Magnesium sulfate?

1 answer:

3 0

Magnesium nitrate o not react with magnesium sulfate because it cannot displace the ions involved. The cations are the same so they just cancel the charges. If you look at it, even of a reaction occurs the product will still be the same substances.

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What are the uses of carbon-14?

Gelneren [198K]

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If the temperature we're to go down how would that effect the solubility of each solute

vova2212 [387]

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3 years ago

Difference between Physical and chemical changes At least six examples for each type of change

Tju [1.3M]

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2 years ago

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6. The graph below shows the heating curve for ethanol (from –200C to 150C). Calculate the amount of heat (kJ) required for each

Kazeer [188]

This problem is providing the heating curve of ethanol showing relevant data such as the initial and final temperature, melting and boiling points, enthalpies of fusion and vaporization and specific heat of solid, liquid and gaseous ethanol, so that the overall heat is required and found to be 1.758 kJ according to:

<h3>Heating curves:</h3>

In chemistry, we widely use heating curves in order to figure out the required heat to take a substance from a temperature to another. This process may involve sensible heat and latent heat, when increasing or decreasing the temperature and changing the phase, respectively.

Thus, since ethanol starts off solid and end up being a vapor, we will find five types of heat, three of them related to the heating-up of ethanol, firstly solid, next liquid and then vapor, and the other two to its fusion and vaporization as shown below:

$Q_T=Q_1+Q_2+Q_3+Q_4+Q_5$

Hence, we begin by calculating each heat as follows, considering 1 g of ethanol is equivalent to 0.0217 mol:

$Q_1=0.0217mol*111.5\frac{J}{mol*\°C}[(-114.1\°C)-(-200\°C)] *\frac{1kJ}{1000J} =0.208kJ\\
\\
Q_2=0.0217mol*4.9\frac{kJ}{mol} =0.106kJ\\
\\
Q_3=0.0217mol*112.4\frac{J}{mol*\°C}[(78.4\°C)-(-114.1\°C)] *\frac{1kJ}{1000J} =0.470kJ\\
\\
Q_4=0.0217mol*38.6\frac{kJ}{mol} =0.838kJ\\
\\
Q_5=0.0217mol*87.5\frac{J}{mol*\°C}[(150\°C)-(78.4\°C)] *\frac{1kJ}{1000J} =0.136kJ$

Finally, we add them up to get the result:

$Q_T=0.208kJ+0.106kJ+0.470kJ+0.838kJ+0.136kJ\\
\\
Q_T=1.758kJ$

Learn more about heating curves: brainly.com/question/10481356

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2 years ago

Explain how the determination of the rate law equation significantly differs from the determination of the equilibrium constant

LUCKY_DIMON [66]

The rate law equation is different from the determination of the equilibrium constant keq expression because rate law equation is used experimentally
And by looking at the equation of the reaction and chemicals used in the reaction equation Keq is determined.
The value of the reaction quotient, when the chemical reaction approaches the state of equilibrium, this is called equilibrium constant.

3 0

3 years ago