<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>
Explanation:
A)
We know, each mole contains 
atoms.
It is given that mass of one oxygen atom is m= 
.
Therefore, mass of one mole of oxygen, 
.
Putting value of n and 
,
B)
Given,
Mass of water in glass=0.050 kg = 50 gm.
From above part mass of one mole of oxygen atoms = 16.0 gm.
Therefore, number of mole of oxygen equivalent to 50 gm oxygen
LEARN MORE :
Avogadro's number
brainly.com/question/12902286
<h2>Hey there!</h2>
<h3>The correct option is (A) It has a partial negative charge on oxygen and a partial positive charge on hydrogen.</h3>
<h3>☆ Explanation:</h3>
¤ As water has the ability to form hydrogen bonds which makes it an excellent solvent.
¤ For this ability of water it can dissolve many different kinds of molecules.
<h2>Hope it helps </h2>
Answer:
The answers are in the explanation.
Explanation:
The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:
Increasing temperature of ice from -10°C - 0°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g
Q = 2.06J/g°C*10°C*10g
Q = 206J
Change from solid to liquid:
The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:
Q = 333.55J/g*10g
Q = 3335.5J
Increasing temperature of liquid water from 0°C - 100°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g
Q = 4.18J/g°C*100°C*10g
Q = 4180J
Change from liquid to gas:
The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:
Q = 2260J/g*10g
Q = 22600J
Increasing temperature of gas water from 100°C - 120°C:
Q = S*ΔT*m
Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g
Q = 1.87J/g°C*20°C*10g
Q = 374J
Total Energy:
206J + 3335.5 J + 4180J + 22600J + 374J =
30695.5J =
30.7kJ
Answer:
+15.8°
Explanation:
The formula for the observed rotation (α) of an optically active sample is
α = [α]<em>lc
</em>
where
<em>l</em> = the cell path length in decimetres
<em>c</em> = the concentration in units of g/100 mL
[α] = the specific rotation in degrees
1. Convert the concentration to units of g/100 mL
2. Calculate the observed rotation
Q6. 3
Q7. 3
Q8. pH
Q18. 3
Q19. 3
Q20. 4
Hope this helped？？
