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3 years ago

what are the moon phases. but be more detailed and don't put in "the phases of the moon" be more detailed.

2 answers:

8 0

As the moon orbits or circles the Earth, the phase changes. We'll start with what is called the New Moon phase. This is where we can't see any of the lit up side of the moon. The moon is between us and the sun (see the picture). As the moon orbits the Earth we can see more and more of the lit up side until finally the moon is on the opposite side of the Earth from the sun and we get a full moon. As the moon continues to orbit the Earth we now see less and less of the lit up side. The phases of the moon starting with the New Moon are: New Moon Waxing Crescent First Quarter Waxing Gibbous Full Waning Gibbous Third Quarter Waning Crescent Dark Moon.

Murrr4er [49]3 years ago

4 0

Explanation:

The lunar phase or Moon phase is the shape of the Moon's directly sunlit portion as viewed from Earth. The lunar phases gradually change over a synodic month as the Moon's orbital positions around Earth and Earth around the Sun shift.

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An atom of which element has the greatest attraction for electrons in a chemical bond?

Katyanochek1 [597]

Answer is (4) - Se.

Among the given choices Se has the highest electronegativity value as 2.4 compared to others. Hence, Se shows greatest attraction for electrons in a chemical bond.

Electronegativity is a value that tells us how an atom can attract electrons towards itself. If the electronegativity is high, then the attraction to the electrons is also high.

5 0

3 years ago

Read 2 more answers

Determine which of the following pairs of reactants will result in a spontaneous reaction at 25°C. None of the above pairs will

Alla [95]

Answer:

Only $Fe^{3+}+Mg$ gives spontaneous reaction.

Explanation:

A redox reaction will be spontaneous if standard reduction potential ( $E^{0}$ ) of the reaction is positive. Because it leads to negative standard gibbs free energy change ( $\Delta G^{0}$ ), which is a thermodynamic condition for spontaneity of a reaction.

$E^{0}=E^{0}(reduction)-E^{0}(oxidation)$

Where $E^{0}(reduction)$ and $E^{0}(oxidation)$ represents standard reduction potential of reduction half cell and standard reduction potential of oxidation half cell.

(1) Oxidation: $Mg-2e^{-}\rightarrow Mg^{2+}$ ; $E_{Mg^{2+}\mid Mg}^{0}=-2.38V$

Reduction: $Fe^{3+}+3e^{-}\rightarrow Fe$ ; $E_{Fe^{3+}\mid Fe}^{0}=-0.04V$

So, $E^{0}=E_{Fe^{3+}\mid Fe}^{0}-E_{Mg^{2+}\mid Mg}^{0}=(-0.04+2.38)V=2.34V$

Hence this pair will give spontaneous reaction.

(2) Similarly as above, $E^{0}=E_{Pb^{2+}\mid Pb}^{0}-E_{Au^{+}\mid Au}^{0}=(-0.13-1.69)V=-1.82 V$

Hence this pair will give non-spontaneous reaction.

(3) Similarly as above, $E^{0}=E_{Ag^{+}\mid Ag}^{0}-E_{Br_{2}\mid Br^{-}}^{0}=(0.80-1.07)V=-0.27 V$

Hence this pair will give non-spontaneous reaction.

(4) Similarly as above, $E^{0}=E_{Li^{+}\mid Li}^{0}-E_{Cr^{3+}\mid Cr}^{0}=(-3.04+0.74)V=-2.30 V$

Hence this pair will give non-spontaneous reaction.

6 0

3 years ago

Why doesn’t the moon move away from earth

Leni [432]

The earth has the moon captured in its gravity. this keeps the moon in an elliptical orbit

8 0

3 years ago

Help pleaseeeeeeeeeee

hichkok12 [17]

Answer:

D 1 and 3 only I am not sure

Explanation:

6 0

3 years ago

Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.

zubka84 [21]

Answer: The value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

Explanation:

For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

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

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

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

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]$

We are given:

$\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol$

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

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

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

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]$

We are given:

$\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol$

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

$\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}$

where,

$\Delta H^o_{rxn}$ = standard enthalpy change of the reaction =-67200 J/mol

$\Delta S^o_{rxn}$ = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol$

Hence, the value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

7 0

3 years ago