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Setler [38]
3 years ago
13

What two positions would the moon be when we have the lowest tidal range?

Chemistry
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer:

Quarter and half moons

Explanation:

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When PCl5 solidifies it forms PCl4+ cations and PCl6– anions. According to valence bond theory,
rewona [7]

Hey there!

No of hybrid orbitals , H = ( V +S - C + A ) / 2

Where H = no . of hybrid orbitals

V = Valence of the central atom = 5

S = No . of single valency atoms = 4

C = No . of cations = 1

A = No . of anions = 0

For PCl4 +

Plug the values we get H = ( 5+4-1+0) / 2

H =  4 ---> sp3 hybridization

sp3 hybrid orbitals are used by phosphorous in the PCl4+ cations

Answer C

Hope that helps!

3 0
3 years ago
Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
zysi [14]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<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:

N_2+2O_2\rightarrow 2NO_2

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

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = \frac{121.77}{1}\times 1.90=231.36 J/K

Hence, the value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0
3 years ago
Salinity is a measure of which of the following in water?
Lemur [1.5K]
The answer is dissolved salts
8 0
3 years ago
Read 2 more answers
If you had excess aluminum, how many moles of aluminum chloride could be produced from 28.0 g of chlorine gas, Cl2?
Mice21 [21]

Answer:

Yes chemistry. Try to add then multiply the top. Get the moles and you will find it.

Explanation:

Try to add then multiply the moles in the equation

6 0
3 years ago
What do you observe about the movement of the particles in the medium?
horrorfan [7]

Answer:

The particles of the medium just vibrate in place.

Explanation:

As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on. Particles of the medium don't actually travel along with the wave.

5 0
3 years ago
Read 2 more answers
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