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3 years ago

7

Suppose the Moon's orbit were not tilted. How many total solar eclipses would you expect to see from Earth in a year?

1 answer:

Pavel [41]3 years ago

5 0

The Moon completes its orbit in about 1 month. If the Moon's orbit were not tilted, a total solar eclipse would occur every time the Moon was between the Earth and Sun, or once a month hope this helps❤️

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31 Which ion has the smallest radius?<br> (1) O2-(2) S2-(3) Se2-<br> (4) Te2-

bearhunter [10]

The answer 1) Oxygen. And another smallest radius is fluorine, because it is the first in the periodic table.

6 0

3 years ago

The first IE of neon (atomic number 10) is significantly higher than that of argon (atomic number 18) but significantly

lbvjy [14]

Explanation:

Different atoms binds their outermost shell electrons with different amount of energy.

The amount of energy required to remove an electron from an atom is the ionization energy.

Ionization energy measures the readiness of an atom to lose electrons.
From the given problem, we can infer that in group O the ionization energy decreases down the group.
Helium has the highest ionization energy.
Down a group on the periodic table, ionization energy decrease because:

atomic radii increases down the group.
there is an increasing shielding/screening effect of inner shell electrons on the outermost shell electrons.

Learn more:

Ionization energy brainly.com/question/2153804

#learnwithBrainly

8 0

2 years ago

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0

2 years ago

The term “periodic” helps identify _______ that occur in the table, like electronegativity or atomic radii.

stepan [7]

Answer: ITS C OR B I THINK BUT I HAVE A FEELING ITS ONE OF THOSE

Explanation:

IM SO SORRY IF ITS WRONG KSDBDZDH

8 0

3 years ago

Read 2 more answers

Find oxidation number for the aucl4 in haucl4 is it 1 or -1?

Romashka [77]

$H^{+I}Au^{+VII}Cl_{4}^{-I}$

7 0

3 years ago