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
Answer:
5
Explanation:
They have 5 in common but different x
Answer:
c = 0.13 j/ g.°C
Explanation:
Given data:
Mass of mercury = 29.5 g
Initial temperature = 32°C
Final temperature = 161°C
Heat absorbed = 499.2 j
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Q = m.c. ΔT
ΔT = T2 - T1
ΔT = 161°C - 32°C
ΔT = 129 °C
Q = m.c. ΔT
c = Q / m. ΔT
c = 499.2 j / 29.5 g. 129 °C
c = 499.2 j / 3805.5 g. °C
c = 0.13 j/ g.°C
Answer:
I can't draw but you could draw 2 electrons in the first orbit and 3 electrons in the second orbit.
Explanation:
This is false. An alcohol does indeed have a polar C-O single bond, but what we should really be focusing on is the extraordinarily polar O-H single bond. When oxygen, fluorine, or nitrogen is bound to a hydrogen atom, there is a small (but not negligible) charge separation, where the eletronegative N, O, or F has a partial negative charge, and the H has a partial positive charge. Water has two O-H single bonds in it (structure is H-O-H). The partially negative charge on the O of the water molecule (specifically around the lone pair) can become attracted either a neighboring water molecule's partially positive H atom, or an alcohol's partially positive H atom. This is weak (and partially covalent) attraction is called a hydrogen bond. This is stronger than a typical dipole-dipole attraction (as would be seen between neighboring C-O single bonds), and much stronger than dispersion forces (between any two atoms). When the solvent (water) and the solute (the alcohol) both exhibit similar intermolecular forces (hydrogen bonding being the most important in this case), they can mix completely in all proportions (i.e. they are miscible) in water.
