Use the formula
first step:
Use the formula
molarity= mole/liter
change ml to l
plug in data
to get .1=mole/.25 or .1M*.25liter
which =.025 moles
then divide .025 moles by two because there are two OH in Sr(OH)2
then multiply that by 265.76 (the molar mass of water)
.0125*265.76
which is 3.32grams this is your answer
The new pressure is larger than the original, the new volume is smaller than 9.0 ml and the new volume is 6.0
good luck :D
Answer:
The relation between the shielding and effective nuclear charge is given as
where s denote shielding
z_{eff} denote effective nuclear charge
Z - atomic number
Explanation:
shielding is referred to as the repulsion of an outermost electron to the pull of electron from valence shell. Higher the electron in valence shell higher will be the shielding effects.
Effective nuclear charge is the amount of net positive charge that valence electron has.
The relation between the shielding and the effective nuclear charge is given as
wheres denote shielding
z_{eff} denote effective nuclear charge
Z - atomic number
After ionization, sodium gains a net positive charge cuz sodium loses its 1 valence electron to gain the nearest stable octet which is neon{Ne}. Hope it helps
Answer:
Explanation has been given below.
Explanation:
- Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
- Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
- First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
- Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
- Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
- Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
