HCl is considered as polar molecule and that is incorrect answer, The choices provided are A) HCl , B) CH₄ , C) H₂O, D) NH₃
And the correct answer is : B) CH₄
In case of H₂O and NH₃ the high difference in electronegativity between nitrogen or oxygen and hydrogen make the bonds acquire partial charges and become polar while in case of CH₄ the difference in electronegativity between carbon and hydrogen is only 0.4 so it considered as non-polar bond
also carbon connected from all directions with Hydrogens
Answer:
Oxygen
Explanation:
Oxygen has 6 valence electrons and it wants to have 8 to obtain a stable octet. So Oxygen gains two electrons in the bond, developing an anion.
It is also a non-metal, and they tend to gain anions.
Answer:
Every element has a proton, neutron, and electron. The number of protons is equal to the atomic number, and the number of electrons is equal to the protons unless it is an ion.
Calcium with water:
When a metal reacts with water( cold water or hot water) then the products formed are metal hydroxide and hydrogen gas.
Metal + steam --------> Metal hydroxide + Hydrogen
Calcium reacts with cold water to form calcium hydroxide and hydrogen gas:
potassium with water.
Potassium reacts violently with cold water to form potassium hydroxide and hydrogen gas:
2K(s)+ 2H2O(l) --------> 2KOH(aq) + H2(g) + heat.
In this reaction so much heat is produced get hydrogen gas formed catches fire and burns explosively.
B. 3.77 L is the new volume occupied by the gas.
<u>Explanation:</u>
As per Avogadro's law, which states that if the pressure and temperature held constant, then an equal volume of the gases will occupy an equal number of molecules. It can be written as,
Here, V1, volume of the helium gas = 2.9 L
V2, volume of the additional helium gas in the balloon = ?
n1, moles of helium gas = 0.150 mol
n2, number of moles of additional helium gas = 0.150 + 0.0450 = 0.195 mol
We have to rearrange the equation for V2 as,
V2 = 
Now Plugin the values as,
V2 = 
= 3.77 L
So the new volume of the balloon is 3.77 L.
