Answer:
The answer would be beryllium
You were probably confused but the correct answer would actually be a beryllium ion.
It's only a small difference (103 degrees versus 104 degrees in water),
and I believe the usual rationalization is that since F is more
electronegative than H, the electrons in the O-F bond spend more time
away from the O (and close to the F) than the electrons in the O-H bond.
That shifts the effective center of the repulsive force between the
bonding pairs away from the O, and hence away from each other. So the
repulsion between the bonding pairs is slightly less, while the
repulsion between the lone pairs on the O is the same -- the result is
the angle between the bonds is a little less.
Hope this helps!
Explanation:
The given data is as follows.
Mass of a lead atom = 
Volume = 2.00 
Density = 11.3 
As it is mentioned that 1 cubic centimeter contains 11.3 grams of lead.
So, in 2 cubic centimeter there will be 
of lead atoms.
One lead atom has a mass of . Therefore, number of atoms present in 22.6 g of lead will be as follows.
= 
Thus, we can conclude that there are of lead are present.
Answer:
P₂ = 2.88 atm
Explanation:
Given data:
Initial volume of gas = 1.8 L
Final volume = 750 mL
Initial pressure = 17.5 Psi
Final pressure = ?
Solution:
We will convert the units first:
Initial pressure = 17.5 /14.696 = 1.2 atm
Final volume = 750 mL ×1L/1000L = 0.75 L
The given problem will be solved through the Boly's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
1.2 atm × 1.8 L = P₂ ×0.75 L
P₂ = 2.16 atm. L/ 0.75 L
P₂ = 2.88 atm
So we have Barium nitrate with a solubility of 8.7g in 100g water at 20°C.
using that relation
i.e.
8.7g (barium nitrate) =100g (water)
1g barium nitrate = 100/8.7 g water
27g barium nitrate = (100/ 8.7 ) × 27
= 310.34 g
therefore,
you need 310.34g of water is in the jar.
