A polar bond is formed with atoms having different electronegativities. The bonding electrons are attracted more towards the atom with greater electronegativity resulting in unequal sharing of electrons. Therefore the molecule develop partial charges and becomes polar. Polar molecules have dipole moment that is the partial charge on molecules due to differences in electronegativity between atoms.
A non-polar bond is formed with atoms having the same electronegativity, hence the bonded pair of electron is shared equally between atoms. Non-polar molecules have no moment.
Note that: symmetrical molecules having polar bonds are non-polar because the dipoles of the bond exert equal and opposite effect. Hence the dipoles cancel the charges.
Example: HCl
In HCl, Cl is more electronegative therefore Cl atom pulls the electron pair of the covalent bond towards itself and develops a partial negative charge. Consequently H develops a partial positive charge. This therfore leads to the formation of a dipole.
Answer:
305 litres of NO gas will be produced from 916 L of NO₂
Explanation:
Given the balanced equation of the chemical reaction as follows:
3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)
Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.
Molar volume of a gas at STP is 22.4 L
Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas
From the mole ratio of NO₂ to NO in the equation of reaction,
Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas
Volume of 13.631 moles of NO gas = 13.631 × 22.4
Volume of NO gas produced = 305.334L
Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L
The tea was no longer hot or (brewed) so the 5th didn’t dissolve like the others because the tea was hot or warm enough anymore it cooled down. So the sugar won’t dissolve no more.
Answer:
4200ml
Explanation:
Converting 3.1kg to g
3.1*1000= 3100g
Since density = mass/volume, then
volume = mass/ density
Therefore volume = 3100/0.74
= 4189.2ml
converting it to two significant figures
= 4200ml
Answer:
The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L
Explanation:
Given data:
Number of moles of HF = 6.62×10⁻³ mol
Volume of HF in litter at STP = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Standard temperature = 273 K
Standard pressure = 1 atm
Now we will put the values in formula.
1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K
V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 148.38×10⁻³ L
Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.