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3 years ago

As a liquid is added to a beaker, the pressure exerted by the liquid on the bottom of the beaker does what

1 answer:

5 0

So if you put a holo ball of paper in the water it sinks and as it sinks it gets smaller cuse pressure so it cuseus things to shrink cuse of pressure

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Consider the following balanced equation: Cu + 2AgNO3 Cu(NO3)2 + 2Ag How many moles of AgNO3 would be needed to produce 2.8 mole

S_A_V [24]

2.8(2/1)=5.6 moles of AgNO3 would be needed

8 0

3 years ago

Read 2 more answers

"Thermite" reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite react

frosja888 [35]

<u>Answer:</u> The given reaction is non-spontaneous in nature.

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$3Mg(s)+Cr_2O_3(s)\rightarrow 3MgO(s)+2Cr(s)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(3\times \Delta S^o_{(MgO(s))})+(2\times \Delta S^o_{(Cr(s))})]-[(3\times \Delta S^o_{(Mg(s))})+(1\times \Delta S^o_{(Cr_2O_3(s))})]$

We are given:

$\Delta S^o_{(Mg(s))}=32.68J/K.mol\\\Delta S^o_{(Cr_2O_3(s))}=81.2J/K.mol\\\Delta S^o_{(MgO(s))}=26.94J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(3\times (26.94))+(2\times (23.77))]-[(3\times (32.68))+(1\times (81.2))]\\\\\Delta S^o_{rxn}=-50.88J/K=-0.0509kJ/K.mol$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

To calculate the standard Gibbs free energy of the reaction, we use the equation:

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy change of the reaction = 665.1 kJ/mol

T = Temperature = 298.15 K

$\Delta S^o$ = standard entropy change of the reaction = -0.0509 kJ/K.mol

Putting values in above equation, we get:

$\Delta G^o=(665.1kJ/mol)-(298.15K\times (-0.0509kJ/K.mol))=680.27kJ/mol$

As, the Gibbs free energy of the reaction is coming out to be positive, the reaction is non-spontaneous in nature.

Hence, the given reaction is non-spontaneous in nature.

4 0

4 years ago

. Monthly measurements of atmospheric CO2 concentration at Mauna Loa began in March 1958. The average CO2 concentration for that

Harrizon [31]

Answer:

$Increase=98.8ppm$

$Average\ increase/year=1.594\frac{ppm}{year}$

Explanation:

Hello,

In this case, since the nowadays concentration of CO2 is 414.51 ppm and the concentration in 1958 was 315.71 ppm, the total increase is computed via the difference between them:

$Increase=414.51ppm-315.71ppm\\\\Increase=98.8ppm$