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3 years ago

What are some unique properties about ice that differ from liquid water to water vapor

2 answers:

7 0

some unique properties about ice that differ from liquid water to water vapor are that it is a solid, not a liquid or gas. It also expands.

katrin2010 [14]3 years ago

3 0

Ice is a substance that forms when a liquid is freeze. when water freezes the sun melts it into liquid and it is also changed into water vapor.

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Convert the speed of light, 3.0 x 10^8 m/s, into mi/hr. (1610 m = 1 mi). Could you please explain step by step thank you!

jenyasd209 [6]

Answer:

$v=6.7\times 10^9\ mi/hr$

Explanation:

Given that,

The speed of light is 3×10⁸ m/s

We need to convert it into mi/hr.

Since, 1 mile = 1610 m

1 hour = 3600 s

So,

$v=3\times 10^8\ \dfrac{m}{s}\\\\=3\times 10^8\ \dfrac{\dfrac{1}{1610}\ \text{miles}}{\dfrac{1}{3600}\ s}\\\\=6.7\times 10^9\ mi/hr$

So, the required speed of light is $6.7\times 10^9\ mi/hr$ .

8 0

3 years ago

German scientist alfred wiener is best known for what hypothesis?

Leona [35]

He was best known for the Continental Drift

3 0

3 years ago

Discuss availability of the lone pair of electrons to electrophile in dative bonding

Free_Kalibri [48]

Answer:

Explanation:

Here, a balance between attraction between nucleus and electrons, and electron-electron, and nuclei-nuclei repulsion play role.

All chemical bonds are formed by overlapping of orbitals. If the electronegativity of the two elements forming the bond is very different (elements from the 1st ,2nd groups with elements of 7th group) then ionic bond are formed. If the electronegativities are more similar, then overlapping is stronger, and covalent bonds are formed.

5 0

3 years ago

CH3-C_=CH + HCl ( 1:2) -->

svlad2 [7]

CH
3

C≡CH
2mole
HCl

A
CH
3

C(Cl)
2

CH
3

Heat
aq.KOH

B
CH
3

COCH
3

5 0

3 years ago

Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2

mina [271]

Answer: The amount of energy released per gram of $B_5H_9$ is -71.92 kJ

Explanation:

For the given chemical reaction:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

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:

$\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]$

Taking the standard enthalpy of formation:

$\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol$